Question

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

Answer 1

`(AP)/(AB)=(2)/(6)=(1)/(3) and (AQ)/(AC)=(3)/(9)=(1)/(3)`
In `Delta APQ and Delta ABC`, we have
`angle A= angle A and (AP)/(AB)=(AQ)/(AC)`
`:. Delta APQ-Delta ABC` [ by SAS- similarity]
`(PQ)/(BC)=(AP)/(AB)=(1)/(3) rArr BC=3PQ`.