Question

In trapezium ABCD, AB is parallel to DC; P and Q are the mid-points of AD and BC respectively. BP produced meets CD produced at point E.

Prove that:

1. Point P bisects BE,
2. PQ is parallel to AB.

Answer 1

The required figure is shown below

(i) From ΔPED and ΔABP,

PD = AP               ...[P is the mid-point of AD]

∠DPE = ∠APB      ....[Opposite angle]

∠PED = ∠PBA      ...[AB || CE]

∴ ΔPED ≅ ΔABP   ...[ASA postulate]

∴ EP = BP

(ii) In Δ ECB,

P is a mid point of BE and

Q is a mid point of BC

∴ PQ || CE   ...(i)  (by mid point theorem)

and CE || AB   ... (ii)

From equation (i) and (ii)

PQ || AB

Hence proved.