5. ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at P. If AB = DC then prove that = a) APAB = PDC b) PA = PD and PC = PB c) AD || BC 165
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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.
Explanation:
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