Question

. In a trapezium ABCD, AB || DC and M is the
midpoint of BC. Through M, a line PQ|| AD
has been drawn which meets AB in P and DC
produced in Q, as shown in the adjoining figure.
Prove that ar(ABCD) = ar(APQD).​

Answer 1

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Answer 2

ar (ABCD)= ar(APQD).

Proved below.

Step-by-step explanation:

Given:

In a trapezium ABCD, AB ||​ DC and M is the mid point of BC (i.e., BM = CM).

Now, in  ∆MPB and ​∆MQC, we have:,

BM = CM     (Given)  

∠BMP = ∠CMQ            (Vertically opposite angles)

∠PBM = ∠QCM           (Alternate angles)

∴​ ∆MPB ≅ ​∆MQC    (ASA congruency)

Also, we know that   congruent triangles have equal area, then

ar (∆MPB)= ar (∆MQC)                    (1)

Now,

ar (ABCD)= ar(APQD)- ar(∆MPB)+ar(∆MQC)

ar (ABCD)= ar(APQD)- ar(∆MQC)+ar(∆MQC)  (from 1 )

ar (ABCD)= ar(APQD).

Hence proved