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  • In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q (Fig.). Prove that ar (ABCD) = ar (APQD)

In trapezium ABCD, AB\parallel DC and L is the mid-point of BC. Through L, a line PQ\parallel AD has been drawn which meets AB in P and DC produced in Q  (Fig.). Prove that ar (ABCD) = ar (APQD)  
 


 

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Solution.

Given: Trapezium ABCD with AB \parallel CD and BL = CL
PQ \parallel AD
To prove: ar(ABCD) = ar(APQD)
Proof: DC is produced at Q and AB \parallel DC
So, DQ \parallel AP
Also, PQ \parallel AD          (given)
Then APQD is a parallelogram
In \triangle CLQ and \triangle BLP
\ CL = BL   [L is mid-point of BC]
\angle LCQ=\angle LBP                             [alternate interior angles]
\angle CQL=\angle LPB                             [alternate interior angle as PQ is a transverse]
\therefore \triangle CLQ\cong \triangle BLP                       [AAS congruence rule]
ar(\triangle CLQ)=ar(\triangle BLP)           [Congruent triangles have equal area] …(i)

 


Now,
ar(ABCD) = ar(APLCD) + ar(\triangle BLP) …(ii)
ar(APQD) = ar(APLCD) + ar(\triangle CLQ) …(iii)
From eq (i), (ii) and (iii)
\Rightarrow ar(ABCD) = ar(APQD)

 

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