Question

ABCD is a cyclic quadrilateral.

Answer 1

Given: $AB\parallel CD$

$AD$ and $BC$ are tangent.

Since $AB\parallel CD$

$\angle BAC=\angle ACD=x$

$\angle ABD=\angle BDC=y$

$\angle APB=\angle DPC=z$

In $△ABP$

$\angle ABP+\angle BPA+\angle PAB={180}^{\circ }$

$⟹x+y+z={180}^{\circ }$

According to alternate segment theorem, angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

Since $AD$ and $BC$ are tangents

$\angle ADP=\angle PCD=y$

$\angle BCP=\angle PDC=x$

$\angle DPC$ is an external angle to $△PBC$

$⟹\angle DPC=\angle PBC+\angle BCP$

$\angle PBC=\angle DPC–BCP=z-x$

$\angle ABC+\angle ADC=(\angle ABP+\angle PBC)+(\angle ADB+\angle BDC)$

$=x+(z–x)+x+y=x+y+z$

WKT

$x+y+z={180}^{\circ }$

$⟹\angle ABC+\angle ADC={180}^{\circ }$

$⟹$ Opposite angles are supplementary.

ABCD is a cyclic quadrilateral.