Question

ABCD is a trapezium in which AB is parallel to DC. O is the mid-point of BC L. Through the point O,a line PQ parallel to AD has been drawn which intersects AB at Q and DC produced at P. prove that ar(ABCD)=ar(AQPD).

Answer 1

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Answer 2

$<b>Question</b>$ : In trapezium ABCD, AB || DC and L is mid point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. Prove that ar(ABCD) = ar(APQD)

$<b>Given </b>$: AB || CD and PQ || AD where L is mid point on BC.

$<b>To Prove </b>$: ar(ABCD) = ar(APQD)

$<b>Proof </b>$: When AB || CD Or QD || AP and PQ || AD then APQD is a parallelogram.

In ∆CLQ and ∆PLB we have,

∠LCQ = ∠LBP [ Alternate angles ]

LC = LB [ L is mid point on BC ]

∠CLQ = ∠PLB [ Vertically opposite angles ]

Hence ∆CLQ and ∆PLB are congruent by ASA congruency.

Therefore, ar(∆CLQ) = ar(∆PLB)

Adding ar(ADCLP) both sides we get,

ar(ADCLP) + ar(∆CLQ) = ar(∆PLB) + ar(ADCLP)

ar(APQD) = ar(ABCD)

$\bold{\large{Q.E.D}}$

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