Question

In the adjoining figure, ABCD is a trapezium in which AB || DC and P, Q are the midpoints of AD and BC respectively. DQ and AB when produced to meet at E. Also, AC and PQ intersect at R. Prove that

(i) DQ = QE,

(ii) PR || AB and

(iii) AR = RC.

Answer 1

(i) In △ QCD and △ QBE

We know that ∠ DQC and ∠ BQE are vertically opposite angles

So we get

∠ DQC = ∠ BQE

From the figure we know that Q is the midpoint of BC

It can be written as

CQ = BQ

We know that AE || DC and BC is a transversal

From the figure we know that ∠ QDC and ∠ QEB are alternate angles

∠ QDC = ∠ QEB

By ASA congruence criterion

△ QCD ≅ △ QBE

DQ = QE (c. p. c. t)

Therefore, it is proved that DQ = QE

(ii) Based on the midpoint theorem

We know that PQ || AE

From the figure we know that AB is a part of AE

So we have PQ || AB

We know that the intercepts on AD are made by the lines AB, PQ and DC

So we get PQ || AB || DC

We know that PR is a part of PQ and is parallel to AB

So we get PR || AB || DC

Therefore, it is proved that PR || AB.

(iii) From the figure we know that the lines PR, AB and DC are cut by AC and AD respectively.

Based on the intercept theorem we know that AR = RC.

Therefore, it is proved that AR = RC.