Question

abcd is a cyclic quadrilateral whose diagonal ac and bd intersect at p if ab = dc then prove that triangle pas is congruent to triangle pdc ,pa=pb ,pc=pb ,ad is parallel to bc

Answer 1

Answer:

Step-by-step explanation:

Since abcd is a cyclic quadrilateral whose diagonal ac and bd intersect at p if ab = dc, our aim is to prove that triangle pas is congruent to triangle pdc ,pa=pb ,pc=pb ,ad is parallel to bc.

Hence, for the fist part, we have:

In order to prove that ΔAPB and ΔDPC are congruent:

∠ABD = ∠ACD

⇔ ∠BAC = ∠BDC

⇔ AB = DC

⇒ ΔAPB ≅ ΔDPC

For the second part we can conclude that PA = PD and PB = PC by C.P.C.T.

In third part, we are given that diagonals are bisectors, so ABCD is a parallelogram, therefore AD║BC.

Answer 2

Answer:

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Step-by-step explanation:

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