Question

In figure, two circles intersect each other at points A and B . Their common tangent PQ touches the circle at points P, and Q as shown.
Prove that , $\angle PAQ$ + $\angle PBQ=180$

Answer 1

Now, $\angle APQ=\angle ABP$ ....$\left(1\right)$ since angle in the alternate segment are equal.

Now, $\angle AQP=\angle ABQ$ ....$\left(2\right)$ since angle in the alternate segment are equal.

In $△PAQ,$

$\angle PAQ+\angle APQ+\angle AQP={180}^{\circ }$ by angle sum property

$\Rightarrow \angle PAQ+(\angle ABP+\angle ABQ)={180}^{\circ }$ from $\left(1\right)$ and $\left(2\right)$

$\Rightarrow \angle PAQ+\angle PBQ={180}^{\circ }$

Hence proved.