Question

In the given figure $P$ and $Q$ are points on sides $AB$ and $AC$ respectively of $△ABC.$ If $AP=3cm,PB=6cm,AQ=5cm$ and $QC=10cm,$ show that $BC=3PQ.$

Answer 1

We have,

$AB=AP+PB$

$=(3+6)cm$

$=9cm$

$AC=AQ+QC$

$=(5+10)cm$

$=15cm$

$\therefore$ $\frac{AP}{AB}=\frac{3}{9}$

$=\frac{1}{3}$

And,

$\frac{AQ}{AC}=\frac{5}{15}$

$=\frac{1}{3}$

$\Rightarrow$ $\frac{AP}{AB}=\frac{AQ}{AC}$

Thus, in $△APQ$ and $△ABC,$ we have

$\frac{AP}{AB}=\frac{AQ}{AC}$ and $\angle A=\angle A$ [Common]

Therefore, by $SAS$-criterion of similarity, we have

$△APQ\sim △ABC$

$\Rightarrow$ $\frac{AP}{AB}=\frac{PQ}{BC}=\frac{AQ}{AC}$

$\Rightarrow$ $\frac{PQ}{BC}=\frac{1}{3}$

$\Rightarrow BC=3PQ$

Hence proved.