plzz solve 32th question step by step
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Step-by-step explanation:
In AECF-----
AE//FC (since AB//CD)
AE=CF (since half of equal sides AB and CD)
Thus, AECF is a //gm.
Also, AF//EC (opp. sides of //gm)
Now, In ∆CDQ-----
FP//CQ (part of AF and EC) and F is mid-point of CD.
Thus, P must be mid-point of DQ by Mid-Point theorem.
Also, DP=QP------(i)
Similarly in ∆ABP,
QP=BQ------(ii)
From (i) & (ii),
DP=PQ=BQ--------(iii)
DP+PQ+BQ= BD
DP+DP+DP= BD {from(iii)}
3DP =BD
DP=1/3BD
May this help u....
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