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Questions and Answers
In parallelogram $ABDC$, derived from parallel lines $m$ and $n$ with perpendicular segments $AC$ and $BD$, what theorem justifies the conclusion that $AC = BD$?
In parallelogram $ABDC$, derived from parallel lines $m$ and $n$ with perpendicular segments $AC$ and $BD$, what theorem justifies the conclusion that $AC = BD$?
- The diagonals of a parallelogram are congruent.
- Opposite sides of a parallelogram are congruent. (correct)
- Adjacent sides of a parallelogram are congruent.
- Opposite angles of a parallelogram are congruent.
If three parallel lines cut congruent segments on one transversal, then they must cut proportional segments on every transversal.
If three parallel lines cut congruent segments on one transversal, then they must cut proportional segments on every transversal.
False (B)
In $\triangle ABC$, if point $M$ is the midpoint of side $AB$ and $MN$ is parallel to $BC$, what is the name of the theorem that allows us to conclude that $N$ is the midpoint of $AC$?
In $\triangle ABC$, if point $M$ is the midpoint of side $AB$ and $MN$ is parallel to $BC$, what is the name of the theorem that allows us to conclude that $N$ is the midpoint of $AC$?
Triangle Midpoint Theorem
A segment that joins two midpoints of a triangle is called a ______.
A segment that joins two midpoints of a triangle is called a ______.
Which of the following statements is true regarding a midsegment of a triangle?
Which of the following statements is true regarding a midsegment of a triangle?
Match the parts of the proof with the corresponding justifications related to the theorem stating that if three parallel lines cut congruent segments on one transversal, then they cut congruent segments on every transversal:
Match the parts of the proof with the corresponding justifications related to the theorem stating that if three parallel lines cut congruent segments on one transversal, then they cut congruent segments on every transversal:
Given parallel lines $AX$, $BY$, and $CZ$, which statement correctly relates the angles formed by the transversals when proving that if three parallel lines cut congruent segments on one transversal, then they cut congruent segments on every transversal?
Given parallel lines $AX$, $BY$, and $CZ$, which statement correctly relates the angles formed by the transversals when proving that if three parallel lines cut congruent segments on one transversal, then they cut congruent segments on every transversal?
If a line passes through the midpoint of one side of a triangle and is perpendicular to another side, it bisects the third side.
If a line passes through the midpoint of one side of a triangle and is perpendicular to another side, it bisects the third side.
Flashcards
Parallel Lines Equidistant
Parallel Lines Equidistant
If lines m and n are parallel, any point on m is the same distance from line n.
Parallel Lines & Transversals
Parallel Lines & Transversals
If parallel lines cut equal segments on one transversal, they do so on any transversal.
Midpoint Parallel Theorem
Midpoint Parallel Theorem
A line from the midpoint of one side, parallel to another, hits the third side's midpoint.
Midsegment
Midsegment
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Triangle Midsegment Theorem (1/2)
Triangle Midsegment Theorem (1/2)
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Triangle Midsegment Theorem (2/2)
Triangle Midsegment Theorem (2/2)
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Equidistant Parallel Lines
Equidistant Parallel Lines
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Parallelogram
Parallelogram
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Study Notes
- Geometry Lesson 7C: More Parallel Line Theorems
Theorem: Equidistant Points on Parallel Lines
- If two lines are parallel, then all points on one line are equidistant from the other line
Given:
- Line m is parallel to line n (m || n)
- Line AC is perpendicular to line n (AC ⊥ n)
- Line BD is perpendicular to line n (BD ⊥ n)
- A and B are any points on m
Prove:
- AC = BD
Proof:
- AB || CD because AB and CD are contained in parallel lines
- AC || BD because both are coplanar and perpendicular to line n
- Quadrilateral ABDC is a parallelogram
- Opposite sides of a parallelogram are congruent; therefore, AC = BD
Theorem: Proportional Segments on Transversals
- If three parallel lines cut congruent segments on one transversal, then they cut congruent segments on every transversal
Given:
- Line AX is parallel to line BY is parallel to line CZ (AX || BY || CZ)
- Segment AB is congruent to segment BC (AB = BC)
Prove:
- Segment XY is congruent to segment YZ (XY = YZ)
Proof:
- Draw XR and YS parallel to AC, creating parallelograms AXRB and BYSC
- Opposite sides of a parallelogram are congruent, so XR ≅ AB and YS ≅ BC
- By the transitive property, XR ≅ YS (since AB ≅ BC)
- Because parallel lines are cut by transversals, ∠1 ≅ ∠3, ∠3 ≅ ∠4, and ∠4 ≅ ∠2
- By the transitive property, ∠1 ≅ ∠2
- ∠5 ≅ ∠6 because they are alternate interior angles on parallel lines
- ΔXYR ≅ ΔYZS by AAS, so XY ≅ YZ by CPCTC
Theorem: Midpoint and Parallel Line in a Triangle
- A line that contains the midpoint of one side of a triangle and is parallel to another side passes through the midpoint of the third side
Given:
- M is the midpoint of segment AB
- Segment MN is parallel to segment BC (MN || BC)
Prove:
- N is the midpoint of AC
Proof:
- Draw AD parallel to MN
- AB, MN, and BC are three parallel lines that cut congruent segments on transversal AB
- These lines will also cut congruent segments on AC
- AN ≅ NC, so N is the midpoint of AC
Midsegment Definition
- Midsegment: a segment that joins the midpoints of a polygon
Theorem: Midsegment of a Triangle
- A midsegment of a triangle is parallel to the third side and half as long
Given:
- MN is a midsegment of â–³ABC
Prove:
- MN || BC
- MN = (1/2)BC
Proof (Part 1):
- There exists only one line through M parallel to BC
- Line must pass through N, the midpoint of the third side, so MN || BC
Proof (Part 2):
- Let L be the midpoint of BC and draw NL
- NL passes through the midpoint of two sides of the triangle, so is parallel to the third side, AB
- So NL || AB and since we've already proven that MN || BC, MNLB is a parallelogram
- Opposite sides of a parallelogram are congruent, MN = BL
- L is the midpoint of BC, so BL = (1/2)BC and therefore MN = (1/2)BC by the transitive property
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