Question

plzz solve 32th question step by step

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Answer 1

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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