Question

Let A be one point of intersection of
two interesting circles with centre o and q
the tangent at A to the two circle meet the circle 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 ΔABC
​

Answer 1

ANSWER

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.

∴OA⊥AC

⇒PQ⊥AC [∵OAQP is a parallelogram

∴OA∥PQ]

⇒PQ is the perpendicular bisector of AC. [∵Q is the centre of the circle]

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

∴BA⊥AQ

∴BA⊥OP [∵AQPO is parallelogram

∴OP∥AQ]

⇒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

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