Question

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.

Answer 1

We have:

$$\begin{gathered} \frac{AP}{AB}=\frac{2}{6}=\frac{1}{3}\mathrm{and}\frac{\mathrm{AQ}}{\mathrm{AC}}=\frac{3}{9}=\frac{1}{3} \\ \Rightarrow \frac{\mathrm{A}\mathrm{P}}{\mathrm{A}\mathrm{B}}=\frac{\mathrm{AQ}}{\mathrm{AC}} \\ \mathrm{In}△APQ\mathrm{and}△ABC,\mathrm{we}\mathrm{have}: \\ \frac{AP}{AB}=\frac{AQ}{AC} \\ \angle A=\angle A \\ \mathrm{Therefore},\mathrm{by}\mathrm{AA}\mathrm{similarity}\mathrm{theorem},\mathrm{we}\mathrm{get}: \\ \mathit{△}APQ\mathit{~}\mathit{△}ABC \\ Hence\mathit{,}\mathit{}\frac{PQ}{BC}\mathit{}=\frac{\mathit{AQ}}{\mathit{AC}}=\frac{1}{3} \\ \Rightarrow \frac{PQ}{BC}=\frac{1}{3} \\ \Rightarrow BC=3PQ \end{gathered}$$
This completes the proof.