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Question 8
In trapezium ABCD, AB || DC and L is the midpoint 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).


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Solution

Given,
In trapezium ABCD, AB || DC, DC produced to Q and L is the midpoint of BC.
BL = CL
Proof:
DC produced to Q and AB || DC.
So, DQ || AB.
In ΔCLQ and ΔBLP,
CL = BL [L is the mid point of BC]
LCQ=LBP [alternate interior angles as BC is a transversal]
CQL=LPB [alternate interior angles as PQ is a transversal]
ΔCLQΔBLP [by AAS congruence rule]
Then, ar(ΔCLQ)=ar(ΔBLP) …(i)
Now, ar(ABCD)=ar(APQD)ar(ΔCQL)+ar(ΔBLP)
=ar(APQD)ar(ΔBLP)+ar(ΔBLP) [from Eq. (i)]
ar (ABCD) = ar (APQD)


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