Question

Abcd is a cyclic quadrilateral whose diagonals ac and bd intersect at p. If o is the centre of the circle and ab=dc, prove that:1. Triangle PAB congruent to triangle PDC2. PA=PD and PC=PB3. AD//BC

Answer 1

Answer:

Step-by-step explanation:

Abcd is a cyclic quadrilateral

so AC = BD = Diameter of Circle

so AC & BD will intersect at O center of circle

That means P & O are same point

PA = PD = radius

PC = PB = Radius

AB = DC given

ΔPAB ≅ Δ PCD

∠DCP = ∠BAP   & ∠CDP = ∠ABD

=> AB ║ DC

Answer 2

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