Question

ABCD is a trapezium in which AB//DC, O is midpoint of BC. Through the point O a line PQ//DA has been drawn which intersects AB at Q and DC produced at P. Prove that ar(ABCD) = ar(AQPD)

Answer 1

Answer:

Step-by-step explanation:

O is the midpoint of BC.

Implies, BO = CO = 1/2 BC  -(1)

DP // AB and BC is the transversal.

Implies, ∠PCO = ∠QBO (Alternate Interior Angles)  -(2)

∠COP = ∠BOQ (Vertically Opposite Angles)  -(3)

From (1), (2) and (3)

ΔCOP ≅ ΔBOQ (By ASA rule)

Implies, ar(ΔCOP) = ar(ΔBOQ)  -(4)

ar(ABCD) = ar(ADCOQ) + ar(BOQ)

                = ar(ADCOQ) + ar(COP)  [from 4]

∴ ar(ABCD) = ar(AQPD)

Hence, proved.

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)

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

$\textbf{please mark as brainliest}$