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.


Answer :

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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