Question

Prove that the tangent drawn from an external point of a circle subtend equal angle at the centre.

Answer 1

A circle with centre O and a point P outside it.

PA and PB are @ tangents to circle

To prove:

$\angle AOP=\angle BOPand,\angle APO=\angle BPO$

Proof:

In $\Delta AOPand\Delta BOP$

$PA=PB$ (tangents drawn from an external points are equal)

$OP=OP$ (common)

$OA=OB$ (radii of same circle )

$\Delta AOP\cong \Delta BOP\left(SSS\right)\therefore \angle AOP=\angle BOPand\angle APO=\angle BPO\left(CPCT\right)$