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
Answers
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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