In triangle $A B C$ , $P Q$ is a line segment intersecting $A B$ at $P$ and $A C$ at $Q$ such that $P Q ∥ B C$ and $P Q$ divides triangle $A B C$ into two parts in equal area Find $B P / A B$ .

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Solution

$\begin{matrix} G i v e n P Q i s p a r a l l e l t o B C a n d P Q d i v i d e s t r a i n g l e A B C i n t o t w o p a r t s T o f i n d \frac{B P}{A B} o o f : I n Δ A P Q a n d Δ A B C ∠ A P Q = ∠ A B C (A s P Q i s p a r a l l e l t o B C) ∠ P A Q = ∠ B A C (C o m m o n a n g l e s) s o Δ A P Q \sim Δ A B C (b y A A s i m i l a r i t y) T h e r e f o r e \frac{a r (Δ A P Q)}{a r (Δ A B C)} = \frac{A P^{2}}{A B^{2}} \Rightarrow \frac{a r (Δ A P Q)}{2 a r (Δ A P Q)} = \frac{A P^{2}}{A B^{2}} \Rightarrow \frac{1}{2} = \frac{A P^{2}}{A B^{2}} \Rightarrow \frac{A P}{A B} - \sqrt{\frac{1}{2}} \Rightarrow \frac{(A B - B P)}{A B} = \frac{1}{\sqrt{2}} \Rightarrow 1 - \frac{B P}{A B} = \frac{1}{\sqrt{2}} \Rightarrow \frac{B P}{A B} = 1 - \frac{1}{\sqrt{2}} ∴ \frac{B P}{A B} = \sqrt{2} - \frac{1}{\sqrt{2}} \end{matrix}$

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