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Concurrent Circles In A Triangle

Problem

It can be shown that a unique circle passes through three given points. In triangle ABC three points $A'$, $B'$, and $C'$ lie on the edges opposite $A$, $B$, and $C$ respectively.

Given that the circle $AB'C'$ intersects circle $BA'C'$ inside the triangle at point $P$, prove that circle $CA'B'$ will be concurrent with point $P$.


Solution

Consider the following diagram.

As $AB'PC'$ is a cyclic quadrilateral $\angle A + \angle B'PC' = 180^o$. Similarly $A'PC'B$ is a cyclic quadrilateral so $\angle B + \angle A'PC' = 180^o$.

Therefore $\angle A + \angle B = 360 - (\angle B'PC' + \angle A'PC') = \angle A'PB'$.

However, in $\Delta ABC$, $\angle A + \angle B = 180 - \angle C$.

Hence $\angle A'PB' = 180 - \angle C$ and we show that $CA'PB'$ is a cyclic quadrilateral and the circle passing through $C$, $A'$, and $B'$ is concurrent at $P$.

This result is known as Miquel's theorem and remains true if the common point is outside the triangle...

Prove that if circle $AB'C'$ intersect circle $BA'C'$ outside the triangle at point $Q$ that circle $CA'B'$ will still be concurrent at this point $Q$.

Problem ID: 321 (14 Apr 2007)     Difficulty: 3 Star

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