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Question
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 (Figure). Prove that ar (ABCD) = ar (APQD)
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Solution
Given: In trapezium ABCD, AB || DC, DC produced in Q and L is the mid-point of BC.
∴ BL = CL
To prove: ar (ABCD) = ar (APQD)
Proof: Since, DC produced in Q and AB || DC.
So, DQ || AB
In ΔCLQ and ΔBLP,
CL = BL ...[Since, 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]
∴ ΔCLQ ≅ ΔBLP ...[By AAS congruence rule]
Then, ar (ΔCLQ) = ar (ΔBLP) [Since, congruent triangles have equal area] ...(i)
Now, ar (ABCD) = ar (APQD) – ar (ΔCQL) + ar (ΔBLP)
= ar (APQD) – ar (ΔBLP) + ar (ΔBLP) ...[From equation (i)]
⇒ ar (ABCD) = ar (APQD)
Hence proved.
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