In the adjacent figure 5.44,

Given AB ∥ DC

M is mid-point of AC and N is mid-point of DB

Given ABCD is a trapezium with AB ∥ DC

P and Q are the mid-points of the diagonals AC and BD respectively

The figure is given below:

To Prove:- MN ∥ AB or DC and

In ΔAB

AB || CD and AC cuts them at A and C, then

∠1 = ∠2 (alternate angles)

Again, from ΔAMR and ΔDMC,

∠1 = ∠2 (alternate angles)

AM = CM (since M is the mid=point of AC)

∠3 = ∠4 (vertically opposite angles)

From ASA congruent rule,

ΔAMR ≅ ΔDMC

Then from CPCT,

AR = CD and MR = DM

Again in ΔDRB, M and N are the mid points of the sides DR and DB,

then PQ || RB

⇒ PQ || AB

⇒ PQ || AB and CD ( ∵ AB ∥ DC)

Hence proved.