Angle between tangents and angle subtended by radii are supplementary Using the result that the length of the tangents draw from an external point to a circle are equal, prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the center.


I can easily prove this by the following:

$\angle OQP = \angle ORP =  90^{\circ}$
So in quadrilateral $OQPR$ $ \angle QPR + \angle QOR = 180^{\circ}$ and therefore these angles are supplementary.

But I have not used what is said in the question. How should I do it?
Hint: Try to use congruence. If you can prove that the two triangles are congruent then it is a kite. Now, you can prove it easily.
To prove:  $BPA + BOA = 180^\circ$

Solution: In quadrilateral $AOBP$, $PBO = 90^\circ$ (as the angle between the radius and the tangent to the circle is $90^\circ$). Similarly, $PAO = 90^\circ$. $PAO + PBO + BOA + BPA = 360^\circ$ (the sum of the angles of a quadrilateral), therefore $90^\circ + 90^\circ + BOA + BPA = 360^\circ$, so (since we have proved that $PAO = 90^\circ$ and $PBO = 90^\circ$) $BOA + BPA = 360^\circ - 180^\circ = 180^\circ$.