Question

ABCD is a trapezium with ABII CD. P and Q are mid points of diagonal AC and BD. Prove that PQI|ABIICD and PQ = 1/2 (AB-DC).
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Answer 1

Answer:

Let ABCD is a trapezium in which AB || CD.

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

We have to prove that:

PQ || AB or CD and

PQ = (AB - CD)/2

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

∠1 = ∠2 (alternate angles)

Again from ΔAPR and ΔDPC,

∠1 = ∠2 (alternate angles)

AP = CP (since P is the mid=point of AC)

∠3 = ∠4 (vertically opposite angles)

From ASA congruent rule,

ΔAPR ≅ ΔDPC

Then from CPCT,

AR = CD and PR = DP

Again in ΔDRB, P and Q are the mid points of the sides DR and DB,

then PQ || RB

=> PQ || AB

=> PQ || AB and CD

Again in ΔDRB, P and Q are the mid points of the sides DR and RB,

then PQ = RB/2

=> PQ = (AB - AR)/2

=> PQ = (AB - CD)/2

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