Let XY be the tangent at P.
TPT: CD is || to XY.
Construction: Join AB.
ABCD is a cyclic quadilateral.
∠BAC + ∠BDC= 180° … (1)
∠BDC = 180° – ∠BAC … (2)
Equating (1) and (2)
We get ∠BDC = ∠PAB
Similarly we get ∠PBA = ∠ACD
As XY is tangent to the circle at ‘P’
∠BPY = ∠PAB (by alternate segment there)
∴ ∠PAB = ∠PDC
∠BPY = ∠PDC
XY is parallel of CD.
Hence proved.