P and Q are points on the sides AB and AC, respectively, of ∆ ABC. If AP = 2 cm, PB = 4 cm,


P and Q are points on the sides AB and AC, respectively, of ∆ ABC. If AP = 2 cm, PB = 4 cm, AQ = 3 cm and QC = 6 cm, show that BC = 3PQ.


We have:

$\frac{A P}{A B}=\frac{2}{6}=\frac{1}{3}$ and $\frac{\mathrm{AQ}}{\mathrm{AC}}=\frac{3}{9}=\frac{1}{3}$

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

In $\triangle A P Q$ and $\triangle A B C$, we have :

$\frac{A P}{A B}=\frac{A Q}{A C}$

$\angle A=\angle A$

Therefore, by AA similarity theorem, we get:

$\triangle A P Q \sim \triangle A B C$

Hence, $\frac{P Q}{B C}=\frac{A Q}{A C}=\frac{1}{3}$

$\Rightarrow \frac{P Q}{B C}=\frac{1}{3}$

$\Rightarrow B C=3 P Q$

This completes the proof.


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