Question

Let $A$ be one point of intersection of two intersecting circle with centres $O$ and $Q$. The tangent at $A$ to the two circles meet the circles again at $B$ and $C$, respectively. Let the point $P$ be located so that $AOPQ$ is a parallelogram. Prove that $P$ is the circumcentre of the triangle $ABC$.

Answer 1

In order to prove that $P$ is the circumcentre of $△ABC$, it is sufficient to show that $P$ is the point of intersection of perpendicular bisectors of the sides of

$△ABC$, i.e. $OP$ and $PQ$ are perpendicular bisectors of sides $AB$ and $AC$ respectively.

Now, $AC$ is tangent at $A$ to the circle with center at $O$ and $OA$ is its radius.

$\therefore OA\perp AC$

$\Rightarrow PQ\perp AC$ [$\because OAQP$ is a parallelogram

$\therefore OA\parallel PQ$]

$\Rightarrow PQ$ is the perpendicular bisector of $AC$. [$\because Q$ is the centre of the circle]

Similarly, $BA$ is the tangent to the circle at $A$ and $AQ$ is its radius, through $A$.

$\therefore BA\perp AQ$

$\therefore BA\perp OP$ [$\because AQPO$ is parallelogram

$\therefore OP\parallel AQ$]

$\Rightarrow OP$ is the perpendicular bisector of $AB$.

Thus, $P$ is the point of intersection of perpendicular bisectors $PQ$ and $PO$ of sides $AC$ and $AB$ respectively

Hence, $P$ is the circumcentre of $△ABC$