Question

ABCD is a trapezium in which AB is parallel to CD. If P and Q are mid points of AD and BC respectively, prove that PQ is parallel to AB and PQ = 1 by 2 (AB +CD).

Answer 1

ABCD is a trapezium in which AB || DC and P, Q are mid points of AD, BC respectively.

Join CP and produce it to meet BA produced at R.

In ΔPDC and ΔPAG,

PD = PA ( P is mid point of AD)

∠CPD = ∠RPC ( Vertically opposite angles)

∠PCD = ∠PRA ( alternate angles) ( DC||AB, DC||RB and CR transversal)

∴ ΔPDC ≅ ΔPAR

CD = RA and PC = PR

In ΔCRB,

P is mid point of CR ( PC = PR proved)

Q is a mid point of BC (given)

∴ By mid point theorem PQ ||AB and PQ = (1/2)RB.

But RB = RA + AB = CD + AB

Hence PQ||AB and PQ = (1/2)( AB + CD).