Question

In $△ABC$, $PQ$ is a line segment intersecting $AB$ at $P$ and $AC$ at $Q$ such that $segPQ\parallel segBC$. If $PQ$ divides $△ABC$ into two equal parts means equal in area, find $\frac{BP}{AB}$

Answer 1

Given: $△ABC$, $PQ\parallel BC$

In $△APQ$ and $△ABC$

$\angle PAQ=\angle BAC$ (Common angle)

$\angle APQ=\angle ABC$ (Corresponding angles)

$\angle AQP=\angle ACB$ (Corresponding angles)

Therefore, $△APQ\sim △ABC$ (AAA rule)

hence, $\frac{A\left(△APQ\right)}{A\left(△ABC\right)}=\frac{A{P}^2}{A{B}^2}$

$\frac{1}{2}=\frac{A{P}^2}{A{B}^2}$

$\frac{1}{\sqrt{2}}-1=\frac{AP}{AB}-1$

$\frac{1-\sqrt{2}}{\sqrt{2}}=\frac{AP-AB}{AB}$

$\frac{\sqrt{2}-1}{\sqrt{2}}=\frac{BP}{AB}$