10th Grade Math Theorems: Similarity & Pythagoras- Part 1

Introduction

• The video covers theorems from Chapters 1 and 2 (Similarity and Pythagoras Theorem) for 10th-grade board exams in 2025.
• This is Part 1 of the video series.
• Naushad from JR College channel is the presenter, who typically handles Math 1 and Math 2.
• The discussed theorems are important for exam preparation.

Basic Proportionality Theorem (BPT)

• Theorem Statement: If a line is parallel to one side of a triangle and intersects the other two sides at distinct points, the sides are divided in the same proportion.
• If line PQ is parallel to side BC of triangle ABC, then AP/PB = AQ/QC.
• Given: In triangle ABC, line PQ is parallel to side BC.
• To Prove: AP/PB = AQ/QC.
• Construction: Join BQ and CP.
• Proof:
  • Area of a triangle = 1/2 * base * height.
  • Triangles with the same height have areas proportional to their bases.
  • Triangles APQ and BPQ have the same height.
  • Area(APQ) / Area(BPQ) = AP/PB given triangles have equal height.
  • Triangles APQ and CPQ have the same height
  • Area(APQ) / Area(CPQ) = AQ/QC given triangles have equal height.
  • Area(BPQ) = Area(CPQ), since the triangles have equal base and equal height.
  • Therefore, AP/PB = AQ/QC.

Angle Bisector Theorem

• Theorem Statement: The bisector of an angle of a triangle divides the side opposite the angle in the ratio of the remaining sides.
• Given: In triangle ABC, BD bisects angle B.
• To Prove: AB/BC = AD/DC
• Construction: Extend AB to E, draw CE parallel to BD, intersecting AE at E.
• Proof:
  • Angle ABD = Angle CBD because BD is the bisector.
  • BD is parallel to CE by construction.
  • Angle BDC = angle DCE.
  • BC = BE because sides opposite to equal angles are equal.
  • In triangle ACE, BD is parallel to CE.
    • Then, AD/DC = AB/BE by BPT.
  • Since BE = BC, AB/BC = AD/DC.

Theorem on Areas of Similar Triangles

• Theorem Statement: The ratio of areas of two similar triangles equals the ratio of the squares of their corresponding sides.
• Given: Triangle ABC ~ triangle PQR.
• To Prove: Area(ABC)/Area(PQR) = (AB^2)/(PQ^2) = (BC^2)/(QR^2) = (AC^2)/(PR^2).
• Construction: Draw altitudes AM and PN on BC and QR.
• Proof:
  • Area(ABC) / Area(PQR) = (1/2 * BC * AM) / (1/2 * QR * PN) = (BC * AM) / (QR * PN).
  • In triangles ABM and PQN:
    • Angle B = Angle Q because they are corresponding angles of similar triangles.
    • Angle AMB = Angle PNQ = 90 degrees.
  • Triangle ABM ~ triangle PQN by AA test.
    • Then, AB/PQ = AM/PN as corresponding sides of similar triangles.
  • Triangle ABC ~ triangle PQR, then AB/PQ = BC/QR = AC/PR.
  • (BC/QR) * (AM/PN) = (BC/QR) * (AB/PQ) = (BC^2)/(QR^2).
  • Therefore, Area(ABC)/Area(PQR) = (AB^2)/(PQ^2) = (BC^2)/(QR^2) = (AC^2)/(PR^2).

Theorem of Geometric Mean

• Theorem Statement: In a right-angled triangle, the perpendicular from the right angle vertex to the hypotenuse is the geometric mean of the hypotenuse segments.
• Given: Right-angled triangle PQR, angle PQR = 90 degrees, QS is perpendicular to PR.
• To Prove: (QS)^2 = PS * SR
• Proof:
  • Triangle RQS ~ triangle QSP ~ triangle PQR
    • QS/PS = RS/QS
  • Therefore, (QS)^2 = PS * SR (Geometric Mean).

Conclusion

• The video presented four geometry theorems.
• The next video will continue discussing theorems with a focus on the chapter on Circles.

Naushad explains important theorems from Chapters 1 & 2 (Similarity and Pythagoras Theorem) for the 10th-grade board exams in 2025. Part 1 covers the Basic Proportionality Theorem (BPT). Key concepts include triangle properties and area ratios.