Question

In the adjoining figure, △ ABC is a triangle and through A, B, C lines are drawn, parallel respectively to BC, CA and AB, intersecting at P, Q and R. Prove that the perimeter of △ PQR is double the perimeter of △ ABC.

Answer 1

We know that AR || BC and AB || RC

From the figure we know that ABCR is a parallelogram

So we get

AR = BC …….. (1)

We know that AQ || BC and QB || AC

From the figure we know that AQBC is a parallelogram

So we get

QA = BC ……… (2)

By adding both the equations we get

AR + QA = BC + BC

We know that AR + QA = QR

So we get

QR = 2BC

It can be written as

BC = QR/2

BC = ½ QR

In the same way

AB = ½ RP and AC = ½ PQ

Perimeter of △ PQR = PQ + QR + RP

It can be written as

Perimeter of △ PQR = 2AC + 2BC + 2AB

By taking 2 as common

Perimeter of △ PQR = 2 (AC + BC + 2AB)

Perimeter of △ PQR = 2 (Perimeter of △ ABC)

Therefore, it is proved that the perimeter of △ PQR is double the perimeter of △ ABC.