P and Q are points on the side AB and AC respectively of a triangle ABC such that PQ is - Maths -

Question

P and Q are points on the side AB and AC respectively of a
triangle ABC such that PQ is parallel to BC and divides triangle
ABC into two parts, equal in area Find PB : AB

Priyanka Kedia · Accepted Answer

Given : PQ||BC and PQ divides triangle ABC into two parts To Find : PB : AB $\Pr o o f : I n Δ A P Q a n d Δ A B C, ∠ A P Q = ∠ A B C (P Q | | B C) ∠ P A Q = ∠ B A C (C o m m o n a n g l e) \Rightarrow Δ A P Q \approx Δ A B C (A A s i m i l a r i t y) T h e r e f o r e, \frac{a r (Δ A P Q)}{a r (Δ A B C)} = \frac{{A P}^{2}}{{A B}^{2}} \Rightarrow \frac{a r (Δ A P Q)}{2 a r (Δ A P Q)} = \frac{{A P}^{2}}{{A B}^{2}} \Rightarrow \frac{1}{2} = \frac{{A P}^{2}}{{A B}^{2}} \Rightarrow \frac{A P}{A B} = \frac{1}{\sqrt{2}} \Rightarrow \frac{A B - B P}{A B} = \frac{1}{\sqrt{2}} \Rightarrow \frac{A B}{A B} - \frac{B P}{A B} = \frac{1}{\sqrt{2}} \Rightarrow 1 - \frac{B P}{A B} = \frac{1}{\sqrt{2}} \Rightarrow \frac{B P}{A B} = 1 - \frac{1}{\sqrt{2}} \Rightarrow \frac{B P}{A B} = \frac{\sqrt{2} - 1}{\sqrt{2}} \Rightarrow B P : A B = (\sqrt{2} - 1) : \sqrt{2}$ $H e n c e, B P : A B i s (\sqrt{2} - 1) : \sqrt{2}$

P and Q are points on the side AB and AC respectively of a triangle ABC such that PQ is parallel to BC and divides triangle ABC into two parts, equal in area. Find PB : AB