Question

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

Answer 1

We have,
AB=AP+PB=(3+6)cm=9 cm
and, AC = AQ + QC =(5+10)cm = 15cm.
$\therefore \frac{AP}{AB}=\frac{3}{9}=\frac{1}{3}and\frac{1}{3}$
$\Rightarrow \frac{AP}{AB}=\frac{AQ}{AC}$

Thus, in triangles 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
$△APQ\sim △ABC$
$\Rightarrow \frac{AP}{AB}=\frac{PQ}{BC}=\frac{AQ}{AC}$
$\Rightarrow \frac{PQ}{BC}=\frac{AQ}{AC}\Rightarrow \frac{PQ}{BC}=\frac{5}{15}$
$\Rightarrow \frac{PQ}{BC}=\frac{1}{3}\Rightarrow BC=3PQ$