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BxMO Team Selection Test, March, 2020

Geometry Difficulty 7.3 National Olympiad, round 2 Prove it Netherlands

On a circle with centre $M$ there are three distinct points $A$ , $B$ , and $C$ such that $|AB| = |BC|$ . The point $D$ lies inside the circle in such a way that $\triangle BCD$ is isosceles. The second intersection point of $AD$ and the circle is called $F$ . Prove that $|FD| = |FM|$ .

Solution

We will prove that $|FD| = |FC|$ and $|FC| = |FM|$ , which proves the statement.

In the cyclic quadrilateral $ABCF$ , we have $\angle BCF = 180^\circ - \angle BAF$ . As $|AB| = |BC| = |BD|$ we also have $\angle BAF = \angle BAD = \angle ADB$ , hence $\angle BDF = 180^\circ - \angle ADB = 180^\circ - \angle BAF$ . We see that $\angle BCF = \angle BDF$ . Moreover, $\angle DFB = \angle AFB$ and $\angle CFB$ are inscribed angles on chords $AB$ and $BC$ of the same length, hence $\angle DFB = \angle CFB$ . Triangles $BCF$ and $BDF$ have two pairs of equal angles; because they also have the side $BF$ in common, they are congruent. We conclude that $|FC| = |FD|$ and $\angle DBF = \angle CBF$ .

The inscribed angle theorem yields $\angle CMF = 2\angle CBF$ . From the equality $\angle DBF = \angle CBF$ we just found, we get that $2\angle CBF = \angle CBD = 60^\circ$ , hence $\angle CMF = 60^\circ$ . Moreover, $|MC| = |MF|$ (radius of the circle), hence $\triangle CMF$ is isosceles with an angle of $60^\circ$ , which yields that the triangle is equilateral. This means that $|FC| = |FM|$ .

This concludes the proof that $|FD| = |FC| = |FM|$ .

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