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.

Rate this question :

In figure 5.41, seg PD is a median of ΔPQR, Point T is the midpoint of seg PD. Produced QT intersects PR at M. Show that

[Hint : draw DN || QM.]

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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.

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