Question

If the line segment PQ is parallel to side AC of triangle ABC and it divides the triangle in to two parts of equal areas. Find the ratio $\frac{AP}{AB}$.

Answer 1

In triangle ABC

PQ II AC

$\therefore \angle BPQ=\angle BAC$

$\angle B=\angle B$

$\Rightarrow \Delta BPQ\sim \Delta BAC$

$\Rightarrow \frac{BP}{AB}=\frac{BQ}{BC}=\frac{PQ}{AC}$

$\Rightarrow \frac{ar\left(BPQ\right)}{AR\left(BAC\right)}=\frac{1}{2}$

$\frac{ar\left(BPQ\right)}{ar\left(BAC\right)}=\frac{B{P}^2}{A{B}^2}$

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

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

$\Rightarrow \frac{AB}{BP}=\sqrt{2}$

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

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

$\Rightarrow \frac{AP}{BP}=\sqrt{2}-1$

LET AP be $(\sqrt{2}-1)x$ and BP be x

$\therefore AB=(\sqrt{2}-1)x+x=(\sqrt{2}-1+1)x=\sqrt{2}x$

$\therefore \frac{AP}{AB}=\frac{(\sqrt{2}-1)x}{\sqrt{2}x}=\frac{\sqrt{2}-1}{\sqrt{2}}$