Q. 9

# In the adjacent figure 5.44, 􀀍ABCD is a trapezium. AB || DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB.

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.

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