# 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. - Mathematics

Sum

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.

#### Solution

We have,

AB = AP + PB = (3 + 6) cm = 9 cm and, AC = AQ + QC = (5 + 10) cm = 15 cm.

\therefore \frac{AP}{AB}=\frac{3}{9}=\frac{1}{3}\text{ and }\frac{AQ}{AC}=\frac{5}{15}=\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 ∠A = ∠A

Therefore, by SAS-criterion of similarity, we have

∆APQ ~ ∆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}

⇒ BC = 3PQ

Concept: Similarity
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