Question

In trapezium ABCD, AB parallel DC and L is midpoint of BC. Through L, a line PQ parallel AD has been drawn which meets AB in P and DC produced in Q. Pove that ar(ABCD) = ar(APQD).

Answer 1

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Answer 2

ar(ABCD) = ar(APQD)

Proved below.

Step-by-step explanation:

Given:

Here ABCD is a trapezium with AB || CD

L is the mid point of BC and PQ || AD

Now, In ∆CLQ and ∆BLP

∠LCQ = ∠LBP        (∵ AB || CDQ and BC is the transversal so, alternate interior opposite angles)

CL = BL                 (∵ L is the mid point of BC)

∠CLQ = ∠BLP           (Vertically opposite angles)

So,  

∆CLQ ≅ ∆BLP        ( by ASA congruency criteria)

Therefore,

Area (∆CLQ) = Area (∆BLP)

⇒ Area (ADCLP) + Area (∆CLQ) = Area (ADCLP) + Area (∆BLP)

 Hence Area (APQD) = Area (ABCD)