Basic Concepts of Geometry PDF

Basic Concepts of Geometry 5 1 Basic Concepts of Geometry (2) We say line is a set of points which can be...

Basic Concepts of Geometry 5 1 Basic Concepts of Geometry (2) We say line is a set of points which can be Points to Remember: extended in both directions infinitely and indicated by arrow heads.  Distance between two points on a number l line: P R Q Distance between two distinct points can be (3) We can name the line in two ways. calculated by subtracting smaller coordinate (i) By using names of any two distinct points on a from the bigger coordinate. line. (Example: Line PQ or line PR or line RQ) X W M L P A B C D E (ii) By using single small alphabet. –4 –3 –2 –1 0 1 2 3 4 5 (Example: Line l) Example: Plane: (1) Co-ordinate of point A is 1 (1) Plane is an undefined term in geometry. Co-ordinate of point D is 4 (2) We say ‘plane’ is a flat surface which extends 4>1 infinitely in all directions. d(A, D) (3) We can name a plane in two ways: = Greater co-ordinate – Smaller co-ordinate (i) By using names of any three = 4–1 E non-collinear points in a plane. P Q \ d(A, D) = 3 Units (Example: Plane PQR) R (2) Co-ordinate of point M is – 2 (ii) By using a single capital letter. (Example : Plane E) Co-ordinate of point X is – 4 –2>–4 Collinear and Non-collinear points: d(X, M) If there exist a line passing through three or = Greater co-ordinate – Smaller co-ordinate more distinct points, then the points are called = – 2 – (– 4) = – 2 + 4 collinear points, otherwise they are called ‘non-collinear points.’ \ d(X, M) = 2 Units Example: Point: (1) (1) Point is an undefined term in geometry. A B C D (2) We say point is a dot made by a sharp pencil In above figure, points A, B, C, D are collinear on a paper. points. (2) (3) It is represented with the help of a single capital Q R alphabet. S The point A is shown as: P In above figure, points P, Q, R, S are non- A collinear points. Note:  Distance and Betweenness: (i) Point is not a figure in geometry, it just shows the position. If P, Q and R are three collinear points and if d(P, Q) + d(Q, R) = d(P, R) then point Q is (ii) Point does not have length, breadth and height. between P and R. Line: (1) Line is an undefined term in geometry. P Q R (5) 6 Master Key Mathematics - II (Geometry) (Std. IX) (iv) d(J, H) MAstEr KEy QUEstioN sEt - 1 solution: PrACtiCE sEt - 1.1 (Textbook Page No. 5) The co-ordinate of point J is –2. The co-ordinate of point H is –1. (1) Find the distances with the help of the number line given below: –1 > –2 Q P K J H O A B C D E d(J, H) –5 –4 –3 –2 –1 0 1 2 3 4 5 = Greater co-ordinate – smaller co-ordinate = –1 – (–2) (i) d(B, E) = –1 + 2 solution: = 1 The co-ordinate of point B is 2. \ d(J, H) = 1 unit The co-ordinate of point E is 5. 5>2 (v) d(K, o) d(B, E) solution: = Greater co-ordinate – smaller co-ordinate The co-ordinate of point K is –3. = 5–2 The co-ordinate of point O is 0. = 3 0 > –3 \ d(B, E) = 3 units d(K, O) = Greater co-ordinate – smaller co-ordinate (ii) d(J, A) = 0 – (–3) solution: = 0+3 The co-ordinate of point J is –2. = 3 The co-ordinate of point A is 1. \ d(K, o) = 3 units 1 > –2 d(J, A) (vi) d(o, E) = Greater co-ordinate – smaller co-ordinate solution: = 1 – (–2) The co-ordinate of point O is 0. = 1+2 The co-ordinate of point E is 5. 5>0 = 3 \ d(J, A) = 3 units d(O, E) = Greater co-ordinate – smaller co-ordinate (iii) d(P, C) = 5–0 solution: = 5 The co-ordinate of point P is –4. \ d(o, E) = 5 units The co-ordinate of point C is 3. (vii) d(P, J) 3 > –4 solution: d(P, C) The co-ordinate of point P is – 4. = Greater co-ordinate – smaller co-ordinate The co-ordinate of point J is – 2. = 3 – (–4) –2>–4 = 3+4 d(P, J) = 7 = Greater co-ordinate – smaller co-ordinate \ d(P, C) = 7 units = – 2 – (– 4) Basic Concepts of Geometry 7 = –2 + 4 (iii) x = –3, y = 7 = 2 solution: \ d(P, J) = 2 units The co-ordinate of point A is –3. The co-ordinate of point B is 7. (viii) d(Q, B) 7 > –3 solution: d(A, B) The co-ordinate of point Q is –5. = Greater co-ordinate – smaller co-ordinate The co-ordinate of point B is 2. = 7 – (–3) 2 > –5 = 7+3 d(Q, B) = 10 = Greater co-ordinate – smaller co-ordinate \ d(A, B) = 10 units = 2 – (–5) = 2+5 (iv) x = –4, y = –5 = 7 solution: \ d(Q, B) = 7 units The co-ordinate of point A is –4. The co-ordinate of point B is –5. (2) if the co-ordinates of A is x and that of B is y, –4>–5 find d(A, B). d(A, B) (i) x = 1, y = 7 = Greater co-ordinate – smaller co-ordinate solution: = – 4 – (–5) The co-ordinate of point A is 1. = –4+5 The co-ordinate of point B is 7. = 1 7>1 \ d(A, B) = 1 unit d(A, B) = Greater co-ordinate – smaller co-ordinate (v) x = –3, y = –6 = 7–1 solution: = 6 The co-ordinate of point A is –3. \ d(A, B) = 6 units The co-ordinate of point B is –6. –3 > –6 (ii) x = 6, y = –2 d(A, B) solution: = Greater co-ordinate – smaller co-ordinate The co-ordinate of point A is 6. = –3 – (–6) The co-ordinate of point B is –2. = –3 + 6 6 > –2 = 3 d(A, B) \ d(A, B) = 3 units = Greater co-ordinate – smaller co-ordinate = 6 – (– 2) (vi) x = 4, y = – 8 = 6+2 solution: = 8 The co-ordinate of point A is 4. \ d(A, B) = 8 units The co-ordinate of point B is – 8. 4>–8 8 Master Key Mathematics - II (Geometry) (Std. IX) d(A, B) \ Points L, M and N are noncollinear points. = Greater co-ordinate – smaller co-ordinate \ Relation of betweenness does not exist. = 4 – (–8) (v) d(X, y) = 15, d(y, Z) = 7, d(X, Z) = 8. = 4+8 solution: = 12 \ d(X, Y) = 15..... (i) \ d(A, B) = 12 units d(Y, Z) + d(X, Z) = 7 + 8 (3) From the information given below, find which \ d(Y, Z) + d(X, Z) = 15..... (ii) of the point is between the other two. if the \ d(X, Y) = d(Y, Z) + d(X, Z) points are not collinear, state so. [From (i) and (ii)] (i) d(P, r) = 7, d(P, Q) = 10, d(Q, r) = 3. \ Points X, Y and Z are collinear points. solution: \ Relation of betweenness exists. d(P, Q) = 10..... (i) X - Z - Y. d(P, R) + d(Q, R) = 7 + 3 (vi) d(D, E) = 5, d(E, F) = 8, d(D, F) = 6. \ d(P, R) + d(Q, R) = 10..... (ii) solution: \ d(P, Q) = d(P, R) + d(Q, R) [From (i) and (ii)] \ d(E, F) = 8..... (i) \ Points P, Q and R are collinear points. d(D, E) + d(D, F) = 5 + 6 \ Relation of betweenness exists. \ d(D, E) + d(D, F) = 11..... (ii) P - R - Q. \ d(E, F) ≠ d(D, E) + d(D, F) [From (i) and (ii)] (ii) d(r, s) = 8, d(s, t) = 6, d(r, t) = 4. \ Points D, E and F are noncollinear points. solution: \ Relation of betweenness does not exist. \ d(R, S) = 8..... (i) (4) on a number line, points A, B and C are such d(S, T) + d(R, T) = 6 + 4 that d(A, C) = 10, d(C, B) = 8 Find d(A, B) \ d(S, T) + d(R, T) = 10..... (ii) considering all possibilities. \ d(R, S) ≠ d(S, T) + d(R, T) [From (i) and (ii)] solution: \ Points R, S and T are non-collinear points. Case (1) A - B - C \ Relation of betweenness does not exist. \ d(A, C) = d(A, B) + d(B, C) (iii) d(A, B) = 16, d(C, A) = 9, d(B, C) = 7. \ 10 = d(A, B) + 8. solution: \ 10 – 8 = d(A, B) \ d(A, B) = 16..... (i) \ d(A, B) = 2 units. d(C, A) + d(B, C) = 9 + 7 Case (2) A - C - B \ d(C, A) + d(B, C) = 16..... (ii) \ d(A, B) = d(A, C) + d(C, B) \ d(A, B) = d(C, A) + d(B, C) [From (i) and (ii)] \ d(A, B) = 10 + 8. \ Points A, B and C are collinear points. \ d(A, B) = 18 units. \ Relation of betweenness exists. Case (3) B - A - C A - C - B. \ d(A, B) + d(A, C) = d(C, B) (iv) d(L, M) = 11, d(M, N) = 12, d(N, L) = 8. \ d(A, B) + 10 = 8 solution: \ d(A, B) = 8 – 10 \ d(M, N) = 12..... (i) \ d(A, B) = – 2 d(L, M) + d(N, L) = 11 + 8 But distance between two points cannot be negative. \ d(L, M) + d(N, L) = 19..... (ii) \ B - A - C is not possible. \ d(M, N) ≠ d(L, M) + d(N, L) [From (i) and (ii)] \ d(A, B) = 2 units or d(A, B) = 18 units Basic Concepts of Geometry 9 (5) Points X, y, Z are collinear such that \ l(RT) = 2.5 + 3.7 d(X, y) = 17, d(Y, Z) = 8, find d(X, Z). \ l(rt) = 6.2 units solution: Case (1) Consider X - Y - Z (iii) if X - y - Z and l(XZ) = 3 7 , l(Xy) = 7 , then l(yZ) = ? \ d(X, Z) = d(X, Y) + d(Y, Z) solution: \ d(X, Z) = 17 + 8. \ d(X, Z) = 25 units. X Y Z Case (2) X - Z - Y X-Y-Z (Given) \ d(X, Y) = d(X, Z) + d(Z, Y) l(XZ) = l(XY) + l(YZ) \ 17 = d(X, Z) + 8. \ 3 7 = 7 + l(YZ) \ 17– 8 = d(X, Z) \ 3 7 – 7 = l(YZ) \ d(X, Z) = 9 units. \ \ l(yZ) = 2 7 units Case (3) Consider Y - X - Z (7) Which figure is formed by three non-collinear \ d(X, Y) + d(X, Z) = d(Y, Z) points? \ 17 + d(X, Z) = 8. solution: \ d(X, Z) = –17 + 8 \ d(X, Z) = –9 units A B But, distance between two points cannot be negative. Y - X - Z is not possible. C \ d(X, Z) = 25 units or d(X, Z) = 9 units \ By joining three non-collinear points, a triangle is formed. (6) Sketch proper figure and write the answers of the following questions. ProBLEMs For PrACtiCE (i) if A - B - C and l(AC) = 11, l(BC) = 6.5 then l(AB) = ? (1) observe the number line and answer the solution: following questions: E D C B A O P Q R S T A B C –5 –4 –3 –2 –1 0 1 2 3 4 5 A-B-C (Given) l(AC) = l(AB) + l(BC) Find: (i) d(Q, T) (ii) d(E, S) \ 11 = l(AB) + 6.5 (iii) d(O, Q) (iv) d(O, D) \ 11 – 6.5 = l(AB) (2) Draw the figures according to the given information and answer the question: \ l(AB) = 4.5 units When A - B - C, l(AC) = 12, l(BC) = 7.5, then (ii) if r - s - t and l(st) = 3.7, l(rs) = 2.5 then l(A B) = ? l(rt) = ? (3) in each of the following decide whether the solution: relation of betweenness exists or not among the points A, B and D. Name the point which R S T lies between the other two. R-S-T (Given) (i) d(A, B) = 5, d(B, D) = 8, d(A, D) = 11 l(RT) = l(RS) + l(ST) (ii) d(A, B) = 5, d(B, D) = 15, d(A, D) = 17 10 Master Key Mathematics - II (Geometry) (Std. IX) (4) if r - s - t, l(st) = 3.75, l(rs) = 2.15, then goes in a particular direction to infinity is called l(rt) = ? Ray. (5) if X - y - Z, l(XZ) = 5 2 , l(Xy) = 2 2 , then Ray AB. l(yZ) = ? A B  Opposite Rays: ANsWErs Two rays having a common origin and lying (1) (i) 3 units (ii) 9 units on the same line are said to be opposite rays. (iii) 2 units (iv) 4 units A O B (2) l(AB) = 4.5 units. (3) (i) Relation of betweenness does not exist. Here, Ray OA and Ray OB have a common origin O and they lie on same line. (ii) Relation of betweenness exists. Point B is between points A and D. \ OA and OB are called opposite rays. (4) l(RT) = 5.9 units  Length of a line segment: (5) l(YZ)= 3 2 units. The distance between the end points of a line segment is called the length of the segment. Length of segment AB is denoted by l(AB) or Points to Remember: AB.  Congruent segments:  Line segment: Two segments are said to be congruent if they The set consisting of points A and B and all the are of the same length i.e. l(AB) = l(CD) then points between A and B is called the segment seg AB @ seg CD. AB and written as ‘seg AB’.  Properties of Congruent segments: A B (1) Reflexivity: Every segment is congruent to itself. seg PQ @ seg PQ. Note: (2) Symmetry: If the first segment is congruent to (1) The points A and B are called end points of the second, then the second is congruent to first. seg AB. If seg PQ @ seg AB then seg AB @ seg PQ. (2) A line segment is a subset of a line. (3) Transitivity: If first segment is congruent  Midpoint of segment: to second segment and second segment is The point M is said to be the midpoint of congruent to third segment then first segment seg AB, if (1) A - M - B (2) AM = MB is congruent to third segment. A M B If seg PQ @ seg AB and seg AB @ seg XY then seg PQ @ seg XY.  Comparison of segment: If AB < CD, then we say that seg AB is smaller than seg CD and denoted by seg AB < seg PrACtiCE sEt - 1.2 (Textbook Page No. 7) CD. (1) the following table shows points on a number A B line and their co-ordinates. Decide whether the pair of segments given below the table are C D congruent or not.  Ray: Point A B C D E A portion of line which starts at a point and Co-ordinate –3 5 2 –7 9 Basic Concepts of Geometry 11 (i) seg DE and seg AB \ l(AD) = 4 units....(ii) solution: \ l(BC) ¹ l(AD) [From (i) and (ii)] The co-ordinate of point D is – 7 and \ seg BC is not congruent to seg AD. the co-ordinate of point E is 9 (iii) seg BE and seg AD 9>–7 d(D, E) solution: = Greater co-ordinate – Smaller co-ordinate The co-ordinate of point B is 5. = 9 – (– 7) The co-ordinate of point E is 9. = 9+7 9>5 \ d(D, E) = 16 d(B, E) \ l(DE) = 16 units....(i) = Greater co-ordinate – Smaller co-ordinate The co-ordinate of point A is – 3. = 9–5 The co-ordinate of point B is 5. = 4 5>–3 \ d(B, E) = 4 units d(A, B) \ l(BE) = 4 units....(i) = Greater co-ordinate – Smaller co-ordinate The co-ordinate of point A is – 3. = 5 – (– 3) The co-ordinate of point D is – 7. = 5+3 –3>–7 \ d(A, B) = 8 units d(A, D) \ l(AB) = 8 units....(ii) = Greater co-ordinate – Smaller co-ordinate \ l(DE) ¹ l(AB) [From (i) and (ii)] = – 3 – (– 7) = –3+7 \ seg DE is not congruent to seg AB. = 4 (ii) seg BC and seg AD \ d(A, D) = 4 units solution: \ l(AD) = 4 units....(ii) The co-ordinate of point B is 5. \ l(BE) = l(AD) [From (i) and (ii)] The co-ordinate of point C is 2. \ seg BE @ seg AD 5>2 (2) Point M is the midpoint of seg AB. if AB = 8 d(B, C) then find the length of AM. = Greater co-ordinate - Smaller co-ordinate solution: = 5–2 M is midpoint of seg AB.... (Given) = 3 1 \ l(AM) = l(AB) \ d(B, C) = 3 units 2 \ l(BC) = 3 units....(i) 1 \ l(AM) = ×8 The co-ordinate of point A is – 3. 2 The co-ordinate of point D is – 7. \ l(AM) = 4 units –3>–7 d(A, D) (3) Point P is the midpoint of seg CD. if CP = 2.5, = Greater co-ordinate – Smaller co-ordinate find l(CD). = – 3 – (– 7) solution: = –3+7 Point P is midpoint of seg CD..... (Given) 1 = 4 \ l(CP) = l(CD) 2 \ d(A, D) = 4 units 12 Master Key Mathematics - II (Geometry) (Std. IX) 1 \ 2.5 = l(CD) (ii) Write a pair of points equidistant from point 2 Q. \ 2.5 × 2 = l(CD) Ans. Point R and point P. \ l(CD) = 5 units (iii) Find d(U, V), d(P, C), d(V, B), d(U, L). solution: (4) if AB = 5 cm, BP = 2 cm and AP = 3.4 cm, d(U, V) compare the segments. The co-ordinate of point U is – 5. solution: The co-ordinate of point V is 5. AB = 5 cm, BP = 2 cm, AP = 3.4 cm 5 > –5 5 > 3.4 > 2 d(U, V) \ l(AB) > l(AP) > l(BP) = Greater co-ordinate – Smaller co-ordinate (5) Write the answers to the following questions = 5 – (–5) with reference to figure. = 5+5 T S R P Q \ d(U, V) = 10 Units (i) Write the name of the opposite ray of ray RP. d(P, C) Ans. Ray RS is opposite of ray RP. The co-ordinate of point P is – 2. (ii) Write the intersection set of ray PQ and ray RP. The co-ordinate of point C is 4. Ans. Intersection set of ray PQ and ray RP is ray PQ. 4>–2 (iii) Write the union set of ray PQ and ray QR. d(P, C) Ans. The union of ray PQ and ray QR is line QR. = Greater co-ordinate – Smaller co-ordinate (iv) State the rays of which seg QR is a subset. = 4 – (– 2) Ans. Seg QR is a subset of ray QR, ray RQ, ray QS = 4+2 and ray QT. \ d(P, C) = 6 Units (v) Write the pair of opposite rays with common end point r. d(V, B) Ans. Ray RP and ray RS. The co-ordinate of point V is 5. (vi) Write any two rays with common end point The co-ordinate of point B is 2. s. 5>2 Ans. Ray ST and ray SR. d(V, B) (vii) Write the intersection set of ray SP and ray = Greater co-ordinate – Smaller co-ordinate st. = 5–2 Ans. The intersection of ray SP and ray ST is point \ d(V, B) = 3 Units S. d(U, L) (6) Answer the questions with the help of The co-ordinate of point U is – 5. figure. The co-ordinate of point L is – 3. R U Q L P A B C V D –3>–5 –6 –4 –2 0 2 4 6 (i) state the points which are equidistant from d(U, L) point B. = Greater co-ordinate – Smaller co-ordinate Ans. (a) Point A and Point C. = – 3 – (– 5) = – 3 + 5 (b) Point P and Point D. \ d(U, L) = 2 Units Basic Concepts of Geometry 13 ProBLEMs For PrACtiCE The part of the statement following ‘If’ is called (1) the co-ordinates of the points on a line are as the antecedent and the part following ‘then’ is follows: called the consequent. Points P Q R S T For example, consider the statement: The diagonals of a rhombus are perpendicular Co-ordinates –3 4 2 –5 9 bisectors of each other. Check whether given pair of segments are The statement can be written in the conditional congruent or not. form as, ‘If the given quadrilateral is a rhombus (i) seg PQ and seg QS. then its diagonals are perpendicular bisectors (ii) seg RS and seg PQ. of each other.’ (iii) seg PS and seg RT. If the antecedent and consequent in a given (2) If P is midpoint of seg AB and AB = 7 cm, find conditional statement are interchanged, the AP. resulting statement is called the converse of the given statement. (3) if Q is midpoint of seg CD and d(C, Q) = 6, find length of seg CD. If a conditional statement is true, its converse is not necessarily true. Study the following (4) if AB = 7 cm, BP = 4 cm, AP = 5.4 cm, compare the segments. examples. Conditional statement: If a quadrilateral is a (5) In the below figure, rhombus then its diagonals are perpendicular l D C A B E bisectors of each other. A, B, C, D and E are the points of line l. Converse: If the diagonals of a quadrilateral are perpendicular bisectors of each other then (i) Name the opposite rays with point A as it is a rhombus. origin. In the above example, the statement and its (ii) Name the ray opposite to ray BE. converse are true. (iii) Find the intersection of ray CE and ray BC. Now consider the following example. Conditional statement: If a number is a prime ANsWErs number then it is even or odd. (1) (i) not congruent (ii) congruent Converse: If a number is even or odd then it is (iii) not congruent a prime number. In this example, the statement is true, but its (2) AP = 3.5 cm converse is false. (3) l(CD) = 12 cm.  Axioms/Postulates: (4) AB > AP > BP These are simple statements which we accept as (5) (i) ray AD and ray AE true and need not be proved. Such statements (ii) ray BD (iii) seg CB are called Axioms or Postulates. Some Euclid’s Postulates are given below: l There are infinite lines passing through a Points to Remember: point:  Conditional statements and converse: In the following figure, lines l, m, n and o passes The statements which can be written in the ‘If- through point P. Similarly, we can draw infinite then’ form are called conditional statements. lines passing through P. 14 Master Key Mathematics - II (Geometry) (Std. IX) n In the above figure, m a and b form a pair of interior angles. a is an acute angle, i.e. Ða < 90º l P b is an acute angle, i.e. Ðb < 90º \ Ða + Ðb < 90º + 90º o \ Ða + Ðb < 180º l There is one and only one line passing \ line l intersects line m at point T. through two distinct points.  theorem: Line l passes through the points A and B. l l A property is supposed to be true if it can be B proved logically. It is then called Theorem. A Proof: l A circle of given radius can be drawn taking The logical argument made to prove a theorem any point as its centre. is called its proof. A circle of radius 3 cm with centre P is drawn.  theorem: l If two lines intersect each other then the vertically 3 cm P opposite angles formed are congruent. A D l All right angles are congruent to each other. In the following figures, ÐB = ÐQ = ÐM = 90º P Q C B A Given: Line AB and line CD intersect each R P other at point P. B C to prove: (1) ÐAPC @ ÐBPD (2) ÐAPD @ ÐBPC L Proof: ÐAPC + ÐAPD = 180º...(i) (Linear pair of angles) N M ÐBPD + ÐAPD = 180º...(ii) l if two interior angles formed on one side of (Linear pair of angles) a transversal of two lines add upto less than two right angles then the lines produced in \ ÐAPC + ÐAPD = ÐBPD + ÐAPD that direction intersect each other. [From (i) and (ii)] n \ ÐAPC = ÐBPD l i.e. ÐAPC @ ÐBPD a Similarly, we can prove, T b ÐAPD @ÐBPC. m Basic Concepts of Geometry 15 PrACtiCE sEt - 1.3 (Textbook Page No. 11) ProBLEMs For PrACtiCE Q.1. Write the following statement in ‘if-then’ (1) Write the following statement in ‘if-then’ form. form: (i) the opposite angles of a parallelogram are (i) All sides of rhombus are congruent. congruent. (ii) In an equilateral triangle all sides are solution: congruent. (2) Write the converse of following theorems: If a quadrilateral is a parallelogram then its opposite angles are congruent. (i) Opposite angles of a cylic quadrilateral are supplementary. (ii) the Diagonals of a rectangle are congruent. (ii) In an Isosceles triangle the angles opposite to solution: equal sides are congruent. If a quadrilateral is a rectangle then its diagonals are congruent. ANsWErs (iii) In an isosceles triangle, the segment joining (1) (i) If the quadrilateral is a rhombus, then all the vertex and the mid point of the base is sides are congruent. perpendicular to the base. (ii) If the triangle is equilateral, then all three sides are congruent. solution: (2) (i) If opposite angles of a quadrilateral are If a triangle is an isosceles then segment joining supplementary, then the quadrilateral is vertex and midpoint of the base is perpendicular cyclic. to the base. (ii) In a triangle, the sides opposite to congruent Q.2. Write converse of the following statements: angles are congruent. (i) The alternate angles formed by two parallel lines and their transversal are congruent. ProBLEM sEt - 1 (Textbook Page No. 11) solution: (1) select the correct alternative from the answers Converse of above statement: of the questions given below: If alternate angles formed by two lines and its (i) How many mid points a segment have? a transversal are congruent then the lines are (A) Only one (B) Two (C) Three (D) Many parallel. Ans. (A) (ii) If a pair of the interior angles made by a (ii) How many points are there in the intersection transversal of two lines are supplementary then of two distinct lines? the lines are parallel. (A) Infinite (B) Two (C) One (D) Not a single solution: Ans. (C) Converse of above statement: (iii) How many lines are determined by three distinct If two parallel lines are intersected by a points? transversal then interior angles so formed are (A) Two (B) Three (C) One or three (D) Six supplementary. Ans. (C) (iii) The diagonals of a rectangle are congruent. (iv) Find d(A, B), if co-ordinates of A and B are – 2 and 5 respectively. solution: (A) –2 (B) 5 (C) 7 (D) 3 Converse of above statement does not exit. Ans. (C) (v) If P - Q - R and d(P, Q) = 2, d(P, R) = 10, then find d(Q, R). (A) 12 (B) 8 (C) 96 (D) 20 16 Master Key Mathematics - II (Geometry) (Std. IX) Ans. (B) d(P, Q) – d(P, R) = d(Q, R) is a false statement. (2) on a number line, co-ordinates of P, Q, r are 3, (3) Co-ordinates of some pairs of points are given – 5 and 6 respectively. State with reason below. Hence find the distance between each whether the following statements are true or pair. false. (i) 3, 6 (i) d(P, Q) + d(Q, R) = d(P, R) solution: (ii) d(P, R) + d(R, Q) = d(P, Q) Let the co-ordinate of point A be 3 (iii) d(R, P) + d(P, Q) = d(R, Q) and the co-ordinate of point B be 6. (iv) d(P, Q) – d(P, R) = d(Q, R) 6>3 solution: d(A, B) = The co-ordinate of P is 3. = Greater co-ordinate – Smaller co-ordinate The co-ordinate of Q is – 5. = 6–3 The co-ordinate of R is 6. = 3 d(P, Q) = 3 – (–5) = 3 + 5 = 8 units \ d(A, B) = 3 d(Q, R) = 6 – (–5) = 6 + 5 = 11 units \ the distance between the given d(P, R) = 6 – 3 = 3 units pair of points is 3 units. (i) d(P, Q) + d(Q, R) = d(P, R) (ii) – 9, – 1 d(P, Q) + d(Q, R) = 8 + 11 = 19 units....(i) solution: d(P, R) = 3 units....(ii) Let the co-ordinate of point A be – 9 \ d(P, Q) + d(Q, R) ¹ d(P, R) and the co-ordinate of point B be – 1. [From (i) and (ii)] –1>–9 \ d (P, Q) + d (Q, R) = d(P, R) is a false d(A, B) statement. = Greater co-ordinate – Smaller co-ordinate (ii) d(P, R) + d(R, Q) = d(P, Q) = – 1 – (– 9) d(P, R) + d(R, Q) = 3 + 11 = 14 units....(i) = –1+9 d(P, Q) = 8 units....(ii) = 8 \ d(A, B) = 8 units \ d(P, R) + d(R, Q) ¹ d(P, Q) [From (i) and (ii)] \ the distance between the given d(P, R) + d(R, Q) = d(P, Q) is false statement. pair of points is 8 units. (iii) d(R, P) + d(P, Q) = d(R, Q) (iii) – 4, 5 d(R, P) + d(P, Q) = 3 + 8 = 11 units....(i) solution: d(R, Q) = 11 units....(ii) Let the co-ordinate of point A be – 4 \ d(R, P) + d(P, Q) = d(R, Q) and the co-ordinate of point B be 5. [From (i) and (ii)] 5>–4 d(R, P) + d(P, Q) = d(R, Q) is a true statement. d(A, B) (iv) d(P, Q) – d(P, R) = d(Q, R) = Greater co-ordinate – Smaller co-ordinate = 5 – (– 4) d(P, Q) – d(P, R) = 8 – 3 = 5 units....(i) = 5+4 d(Q, R) = 11 units....(ii) = 9 \ d(P, Q) – d(P, R) ¹ d(Q, R) \ d(A, B) = 9 units [From (i) and (ii)] \ the distance between the given pair of points is 9 units. Basic Concepts of Geometry 17 (iv) 0, – 2 (vii) 80, – 85 solution: solution: Let the co-ordinate of point A be 0 Let the co-ordinate of point A be 80 and the co-ordinate of point B be – 2. and the co-ordinate of point B be – 85 0>–2 d(A, B) = Greater co-ordinate – Smaller co-ordinate d(A, B) = 80 – (– 85) = Greater co-ordinate – Smaller co-ordinate = 80 + 85 = 0 – (– 2) = 165 \ 0+2 \ d(A, B) = 165 units \ d(A, B) = 2 \ the distance between the given pair of points is 165 units. \ the distance between the given pair of points is 2 units. (4) Co-ordinate of point P, on a number line is –7. Find the co-ordinates of points on the number (v) x + 3, x – 3 line which are at a distance of 8 units from solution: point P. Let the co-ordinate of point A be x + 3 solution: and the co-ordinate of point B be x – 3 Let co-ordinate of Q be x be a point on positive (x + 3) > (x – 3) side of point P. d(A, B) Co-ordinate of P is – 7. = Greater co-ordinate – Smaller co-ordinate x>–7 = (x + 3) – (x – 3) d(P, Q) = x – (– 7) = x+3–x+3 \ 8=x+7 \ x=8– 7 = 6 \ x=1 \ d(A, B) = 6 units \ Co-ordinate of point Q is 1. \ the distance between the given pair of points is 6 units. Let co-ordinate of ponit R be y be a point on negative side of point P. (vi) – 25, – 47 Co-ordinate of point P is – 7 solution: –7>y Let the co-ordinate of A point be – 25 \ d(P, R) = – 7 – y and the co-ordinate of B point be – 47 \ 8=–7–y d(A, B) \ 8+7=–y = Greater co-ordinate – Smaller co-ordinate \ – y = 15 \ y = – 15 = – 25 – (– 47) = – 25 + 47 \ Co-ordinate of point r is – 15 = 22 (5) Answer the following questions. \ d(A, B) = 22 units (i) If A - B - C and d(A, C) = 17, d(B, C) = 6.5 then d(A, B) = ? \ the distance between the given solution: pair of points is 22 units. A-B-C (Given) 18 Master Key Mathematics - II (Geometry) (Std. IX) \ d(A, C) = d(A, B) + d(B, C) (iii)A triangle is a figure formed by three \ 17 = d(A, B) + 6.5 segments. \ 17 – 6.5 = d(A, B) Ans. If a polygon is three-sided closed figure, then it is a triangle. \ d(A, B) = 10.5 units (iv) A number having only two divisors is called a (ii) If P - Q - R and d(P, Q) = 3.4, d(Q, R) = 5.7, then prime number. find d(P, R) = ? Ans. If a number is a prime number then it has only solution: two divisors. P-Q-R (Given) (8) Write the converse of each of the following \ d(P, R) = d(P, Q) + d(Q, R) statements. \ = 3.4 + 5.7 (i) If the sum of measures of angles in a figure is \ = 9.1 units 180º, then the figure is a triangle. \ d(P, r) = 9.1 units Ans. Converse: If a figure is a triangle then sum of (6) Co-ordinate of point A on a number line is all angles of this figure is 180º. 1. What are the co-ordinates of points on the (ii) If the sum of measures of two angles is 90º then number line which are at a distance of 7 units they are complement of each other. from A? Ans. Converse: If two angles are complementary then solution: their sum is 90º. Co-ordinate of point A is 1. (iii) If the corresponding angles formed by a Let co-ordinate of point B be x be a point on transversal of two lines are congruent then the positive side of point A. two lines are parallel. x>1 Ans. Converse: If two parallel lines are intersected \ d(A, B) = x – 1 by a transversal then the pair of corresponding \ 7=x–1 angles is congruent. \ 7+1=x (iv) If the sum of the digits of a number is divisible \ x=8 by 3 then the number is divisible by 3. Ans. Converse: If a number is divisible by 3 then sum \ Co-ordinate of point B is 8. of digits of this number is divisible by 3. Let co-ordinate of point C be y be a point on negative side of point A. (9) Write the antecedent (given part) and the 1>y consequent (part to be proved) in the following statements. \ d(A, C) = 1 – y (i) If all sides of a triangle are congruent then its all \ 7=1–y angles are congruent. \ y=1–7 Ans. Given: In DABC, side AB @ side BC @ side AC. \ y=–6 to prove: ∠A @ ∠B @ ∠C. \ Co-ordinate of point C is – 6. (ii) The diagonals of a parallelogram bisect each (7) Write the following statements in conditional other. form: Ans. Given: (1) oPQRS is a parallelogram. (i) Every rhombus is a square. (2) Diagonals PR and QS intersect at point M. Ans. If a quadrilateral is a square then it is a rhombus. to prove: (1) PM = RM (2) QM = SM. (ii) Angles in a linear pair are supplementary. (10) Draw a labelled figure showing information Ans. If adjacent angles are supplementary, then they in each of the following statements and write form a linear pair. the antecedent and the consequent. Basic Concepts of Geometry 19 (i) Two equilateral triangles are similar. (3) With the help of the given figure, which of the Ans. A P following statement is true? P Q R Q R (A) d(P, R) + d(R, Q) = d(P, Q) B C (B) d(P, R) + d(P, Q) = d(R, Q) Given: (1) In DABC, side AB @ side BC @ side AC. (C) d(Q, R) – d(P, Q) = d(P, R) (2) In DPQR, side PQ @ side QR @ side PR. to prove: DABC ~ DPQR (D) d(P, R) – d(R, Q) = d(P, Q) (ii) If angles in a linear pair are congruent then each (4) If d(P, R) = 7, d(Q, R) = 3 and d(P, Q) = 4, what of them is a right angle. can we say about points P, Q and R? Ans. (A) P – R – Q (B) Q – P – R A (C) P – Q – R (D) R – P – Q (5) A statement which is universally true and need D B C not be proved is called.................. Given: (1) ∠ABC and ∠ABD form a linear pair. (A) Axiom (B) Postulate (2) ∠ABC @ ∠ABD. (C) both a and b (D) Theorem to prove: ∠ABC = ∠ABD = 90º (6) From a single point, we can draw................. (iii) If the altitudes drawn on two sides of a lines. triangle are congruent then those two sides are (A) One (B) Infinite congruent. (C) Two (D) Five Ans. A (7) Two different points determine................. line. (A) Six (B) Infinite N M (C) Two (D) One and only one B C (8) The co-ordinate of point E is.................. Given: In DABC F C G B H A K D I E J l (1) seg BM ^ side AC, A - M - C. –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 (2) seg CN ^ side AB, A - N - B. (A) – 4 (B) 1 (3) seg BM @ seg CN. (C) 2 (D) – 5 to prove: Side AB@ side AC. (9) The co-ordinate of point J is.................. MCQ’s F C G B H A K D I E J l –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 (1) The number associated with the point is called................. of that point. (A) – 7 (B) 2 (A) Origin (B) Midpoint (C) 0 (D) – 5 (C) Co-ordinate (D) Graph (10) Which points are at a distance of 3 units from (2) With the help of the given number line, what is point A? d(P, T)? F C G B H A K D I E J l D C P Q R S T M N A B –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 (A) Points C, E (B) Points G, I (A) 3 units (B) 4 units (C) Points D, B (D) Points H, E (C) – 3 units (D) – 4 units 20 Master Key Mathematics - II (Geometry) (Std. IX) (11) On the given number line, d(K, J) =.................. (18) In the adjoining figure, angles are in linear pair. F C G B H A K D I E J What is the value of x? l –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 (4x + 60)º (x + 20)º (A) 3 units (B) 2 units (C) 4 units (D) 5 units (12) If P – Q – R, l(PR) = 7 units, l(PQ) = 4 units, (A) 40 (B) 80 l(QR) =.................. (C) 20 (D) 30 (A) 11 units (B) 3 units (19) Which of the following are pair of linear pair (C) 6 units (D) 5 units angles. (13) If AB = 10 units, AC = 7 units, BC = 3 units then A which of the following is correct? S (A) A-B-C D (B) C-B-A P Q R B C (C) A-C-B (D) Point A, B, C are non collinear. (A) (B) (14) If the co-ordinates of points P and Q are 3 and – 5 P L M T respectively then d(P, Q) =.................. (A) 9 units (B) 2 units (C) 8 units (D) 7 units N Q S (15) If d(P, Q) = 10 units, d(P, R) = 18 units and P-Q-R R then d(Q, R) =.................. (C) (D) (A) 28 units (B) 10 units (A) A and B (B) B and D (C) 18 units (D) 8 units (C) C and D (D) A and C (16) If the co-ordinates of points P and Q are 2 (20) AB = CD, CD = EF then.................. and – 2 respectively then d(P, Q) =............... (A) AB = EF, by property of symmetry (A) 2 units (B) 3 2 units (B) AB = EF, by property of reflexivity (C) 2 2 units (D) 4 2 units (C) AB = EF, by property of transitivity (17) What is mÐPQS? (D) AB ¹ EF S P ANsWErs ? 100º Q (1) (C) (2) (B) (3) (D) (4) (C) (5) (C) (6) (B) (7) (D) (8) (B) R (9) (B) (10) (B) (11) (C) (12) (B) (A) 70º (B) 90º (13) (C) (14) (C) (15) (D) (16) (C) (C) 80º (D) 60º (17) (C) (18) (C) (19) (D) (20) (C) Basic Concepts of Geometry 21 ASSIGNMENT - 1 Time : 1 Hr. Marks : 20 Q.1. Observe the number line and answer the following: (4) F P N M L O A B C D E –5 –4 –3 –2 –1 0 1 2 3 4 5 Find (1) d(M, C) (2) d(L, P) (3) d(O, N) (4) d(A, E) Q.2. Solve the following: (10) (1) The points A, B, C are on a line such that d(A, C) = 10, d(C, B) = 8 then find d(A, B). (2) Draw the figure, according to the given information and answer the question when A-B-C, l(AC) = 11, l(BC) = 6.5 then l(AB) = ? (3) P is midpoint of seg AB. If AB is 15, then find length of AP. (4) Write the given statement in ‘if-then’ form for prime numbers are divisible by 1 and itself. (5) Diagonals of a parallelogram bisect each other. Write the proof for the given statement. Q.3. Solve the following: (6) (1) Prove: If two lines intersect each other, then vertically opposite angles are congruent. (2) Co-ordinate of a point P on a number line is – 7. Find the co-ordinates of points which are at a distance of 8 units from point P. vvv 22 Master Key Mathematics - II (Geometry) (Std. IX) 2 Parallel Lines Theorem - 1 : Interior angles theorem Points to Remember:  Theorem : If two parallel lines are intersected by  Parallel Lines a transversal, the interior angles on either side Parallel Lines : The lines which are l of the transversal are supplementary. Given : line l line m and n coplanar and do not intersect each m other are called parallel lines. line n is their d a m In the above figure line l is parallel to line m transversal. c b l Hence as shown Symbolically, it is written as 'line l line m' in the figure ∠a, Transversal : A tranversal is a line that intersects ∠b are interior angles formed on one two or more lines in distinct points. side and ∠c, ∠d are interior angles n In the adjoining figure, formed on other side of the transversal. line n is the transversal To prove : ∠a + ∠b = 180° a d b c for lines l and m. The l ∠d + ∠c = 180° e h two lines and the Proof : Three possibilities arise regarding the f g m transversal determine eight angles at the point sum of measures of ∠a and ∠b. of intersection. (i) ∠a + ∠b < 180° Pairs of Corresponding angles : (ii) ∠a + ∠b > 180° (i) ∠b and ∠f (ii) ∠c and ∠g (iii) ∠a + ∠b = 180° (iii) ∠a and ∠e (iv) ∠d and ∠h Let us assume that the possibility Pairs of Alternate Interior angles : (i) ∠a + ∠b < 180° is true. (i) ∠b and ∠h (ii) ∠c and ∠e Then according to Euclid's postulate, if Pairs of Alternate Exterior angles : the line l and line m are produced will intersect each other or the side of the (i) ∠a and ∠g (ii) ∠d and ∠f transversal where ∠a and ∠b are formed. Pairs of Interior angles : But line l and line m are parallel lines (i) ∠b and ∠e (ii) ∠c and ∠h...(Given)  Important Properties ∴ ∠a + ∠b < 180° impossible...(i) (1) When two lines intersect, the pairs of opposite Now let us suppose that ∠a + ∠b > 180° angles formed are congruent. is true. (2) The angles in a linear pair are supplementry ∴ ∠a + ∠b > 180° (3) When one pair of corresponding angles is But ∠a + ∠d = 180° congruent, then all the remaining pairs of and ∠c + ∠b = 180° corresponding angles are congruent....(Angles in linear pair) (4) Whe one pair of alternate angles is congruent, ∴ ∠a + ∠d + ∠b + ∠c = 180° + 180° = 360° then all the remaining pairs of alternate angles ∴ ∠c + ∠d = 360° – (∠a + ∠b) are congruent. If ∠a + ∠b > 180° then (5) When one pair of interior angles on one side of [360° – (∠a + ∠b)] < 180° the transversal is supplementary, then the other ∴ ∠c + ∠d < 180° pair of interior angles is also supplementary. (22) Parallel Lines 23 ∴ In that case line l and line m produced But, ∠c + ∠b = 180° will intersect each other on the same side...(ii) (Interior angles theorem) of the transversal where ∠c and ∠d are ∴ ∠d + ∠c = ∠c + ∠b...[From (i) and (ii)] formed. ∴ ∠d = ∠b ∴ ∠c + ∠d < 180° is impossible ∴ That is ∠a + ∠b > 180° is impossible MASTER KEY QUESTION SET - 2...(ii) ∴ the remaining possibility, PRACTICE SET - 2.1 (Textbook Page No. 17) ∴ ∠a + ∠b = 180° is true (1) In the given figure, D...[From (i) and (ii)] line RP line MS and ∴ ∠a + ∠b = 180° R 85° P line DK is their H Similarily, ∠c + ∠d = 180° transversal, Note that, in this proof, because of ∠DHP = 85°. M G S the contradictions we have denied Find the measure of K the possibilities ∠a + ∠b > 180° and following angles : ∠a + ∠b < 180° (i) ∠RHD (ii) ∠PHG (iii) ∠HGS (iv) ∠MGK Theorem - 2 : Corresponding angles theorem Solution : Statement : The corresponding angles formed by a (i) ∠DHP + ∠RHD = 180° transversal of two parallel lines are of equal...(Angles in linear pair) measures. n a ∴ 85 + ∠RHD = 180 Given : line l line m l c ∴ ∠RHD = 180 – 85 and line n is a b m ∴ ∠RHD = 95° transversal. To prove : ∠a = ∠b (ii) ∠PHG = ∠RHD...(Vertically opposite angles) Proof : ∠a + ∠c = 180° ∴ ∠PHG = 95°...(i) (Angles in linear pair) But, ∠b + ∠c = 180° (iii) line RP line MS...(Given)...(ii) (Interior angles theorem) On transversal DK, ∴ ∠a + ∠c = ∠b + ∠c...[From (i) and (ii)] ∠DHP = ∠HGS ∴ ∠a = ∠b...(Corresponding angles theorem) ∴ ∠HGS = 85° Theorem - 3 : Alternate angles theorem (iv) ∠HGS = ∠MGK...(Vertically opposite angles) Statement : The alternate angles formed by a ∴ ∠MGK = 85° transversal of two parallel lines are of equal measures. (2) In the given figure, line p line q and line l n and line m are transversals. Measures of some Given : line l line m l angles are shown. Hence find the measures of and line n is a d c b ∠a , ∠b, ∠c, ∠d. p transversal. m q a To prove : ∠d = ∠b 110° b e Proof : ∠d + ∠c = 180° l c 115°...(i) (Angles in linear pair) d m 24 Master Key Mathematics - II (Geometry) (Std. IX) Solution : (4) In the given figure, P ∠a + 110 = 180...(Angles in linear pair) sides of ∠PQR and X ∴ ∠a = 180 – 110 ∠XYZ are parallel to each other. Prove ∴ ∠a = 70° that ∠PQR ≅ ∠XYZ. Y Z Name an angle as 'e' as shown in the figure, Construction : Extend ray Q line p line q...(Given) M R XY to intersect ray QR at On transversal l, point M, such that Q-M-R ∠e = ∠a...(Corresponding angles theorem) Proof : line PQ line XY...(Given) ∴ ∠e = 70 i.e. line PQ line XM...(X - Y - M) ∴ ∠b = ∠e...(Vertically opposite angles) On transversal QR, ∴ ∠b = 70° ∠PQR ≅ ∠XMR...(i) (Corresponding angles theorem) line p line q...(Given) line YZ line QR...(Given) On transversal m, On transversal XM, ∠c = 115° ∠XYZ ≅ ∠XMR...(ii)...(Corresponding angles theorem) (Corresponding angles theorem) ∠d + 115 = 180...(Angles in linear pair) ∠PQR ≅ ∠XYZ...[From (i) and (ii)] ∴ ∠d = 180 – 115 (5) In the given figure, P ∴ ∠d = 65° line AB line CD A n p R B (3) In the given and line PQ is 105° figure, line l 45° transversal. Measure T l line m and d of one of the angles C D line n line a m is given. Hence find Q c b p. Find ∠a, the measures of the following angles : ∠b, ∠c from (i) ∠ART (ii) ∠CTQ (iii) ∠DTQ (iv) ∠PRB. the given measure of an angle. Solution : Solution : (i) ∠BRT + ∠ART = 180° Name an angle as 'd' as shown in the figure,...(Angles in linear pair) ∠d = 45°...(Vertically opposite angles) ∴ 105 + ∠ART = 180 line l line m...(Given) ∴ ∠ART = 180 – 105 On transversal p, ∴ ∠ART = 75° ∠d + ∠a = 180°...(Interior angles theorem) (ii) line AB line CD...(Given) ∴ 45 + ∠a = 180 On transversal PQ, ∴ ∠a = 180 – 45 ∠ART = ∠CTQ ∴ ∠a = 135°...(Corresponding angles theorem) ∠b = ∠a...(Vertically opposite angles) ∴ ∠CTQ = 75° ∴ ∠b = 135° (iii) line AB line CD...(Given) line n line p On transversal PQ, On transversal m, ∠BRT = ∠DTQ...(Corresponding angles theorem) ∠c = ∠b...(Corresponding angles theorem) ∴ ∠DTQ = 105° ∴ ∠c = 135° (iv) ∠PRB = ∠ART...(Vertically opposite angles) ∴ ∠PRB = 75° Parallel Lines 25 PROBLEMS FOR PRACTICE Points to Remember: (1) If AB CD, then find ∠PCD and A 56° B Theorem - 4 : Angle sum property of a triangle ∠CPD from the 100° P Statement : The sum of the measures of all angles adjoining figure. of a triagle is 180°. D A E C D (2) In the adjoining A 65° B figure, measures 35° B C of two angles are y Given : ABC is any triangle given. If x z line ED seg AB E C D To prove : ∠BAC + ∠ABC + ∠ACB = 180° and E - C - D, then Construction : Through point A, draw a find the values of x, y and z. line DE side BC, such that D - A - E. Proof : line DE side BC (3) In given figure, AB DE, ∠ABC = 75° and on transversal AB, ∠CDE = 145° then find ∠BCD. D E ∠DAB = ∠ABC...(i) B 145° A (Alternate angles theorem) 75° on transversal AC, C ∠EAC = ∠ACB...(ii) (4) If AB CD, m∠APQ = 50° and m∠PRD = 127° (Alternate angles theorem) find x an y. P ∠DAB + ∠EAC = ∠ABC + ∠ACB A 50° y B...[Adding equations (i) and (ii)] 127° Adding ∠BAC on both sides, x Q C R D ∴ ∠DAB + ∠EAC + ∠BAC = ∠ABC + E ∠ACB + ∠BAC (5) If AB CD, EF ⊥ CD and ∠GED = 126°. ∴ ∠DAB + ∠EAB = ∠ABC + ∠ACB + Find ∠AGE, ∠GEF, ∠FGE ∠BAC...(Adding addition property) G F ∴ ∠ABC + ∠ACB + ∠BAC = ∠DAB + A B ∠EAB ∴ ∠ABC + ∠ACB + ∠BAC = 180°...(Angles in linear pair) C E D Theorem - 5 : Interior angles test ANSWERS Statement : If the X (1) ∠CPD = 80°, ∠PCD = 44° interior angles formed A P B (2) x = 65°, y = 80°, z = 35° by a transversal of two distinct lines are Q (3) ∠BCD = 40° C D supplementary, then (4) x = 50°, y = 77° t h e t w o l i n e s a re Y (5) ∠AGE = 126°, ∠GEF = 36°, ∠FGE = 54° parallel. Given : A transversal XY intersects lines AB and CD such that ∠BPQ + ∠PQD = 180° To prove : line AB side CD 26 Master Key Mathematics - II (Geometry) (Std. IX) Proof : (Indirect method) Theorem - 7 : Corresponding Angles Test Let us assume line AB is not parallel to Statement : If a pair of n line CD. corresponding angles a ∴ They intersect each other at point T. formed by a transversal of c l A X b two lines is congruent m P D T then the two lines are parallel. B C Q Given : line n is a transversal for lines l and m ∠a = ∠b Y To Prove : lines l line m In PQT, Proof : ∠a + ∠c = 180°...(i) ∠TPQ + ∠PQT + ∠PTQ = 180° (Angles in linear pair)...(Angle sum property of a triangle) But, ∠a = ∠b...(ii) (Given) ∴ ∠BPQ + ∠PQD + ∠PTQ = 180°...(i) [P - T - B, Q - T - D] ∴ ∠b + ∠c = 180° [From (i) and (ii)] But ∠BPQ + ∠PQD = 180°...(ii) (Given) ∴ line l line m...(By Interior angles test) ∴ ∠BPQ + ∠PQD + ∠PTQ = ∠BPQ + ∠PQD...[From (i) and (ii)] Corollary 1 : E ∴ ∠PTQ = 0° Statement : If a line is A B ∴ lines AB and CD are not two distinct perpendicular to two L lines lines in a plane, then the But, this contradicts the given that two two lines are parallel to C M D each other. F lines AB and CD are distinct. ∴ Our assumption that line AB is not Given : line EF ⊥ line AB parallel to line CD is false. line EF ⊥ line