Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.8 National Olympiad Prove it Ireland

Triangle $ABC$ is obtuse with $\angle ABC > 90^\circ$ . Let $\ell$ be the external bisector of $\angle BCA$ , and let $AB$ (extended) meet $\ell$ at $D$ . The circumcircle $\Gamma$ of triangle $BCD$ has centre $O$ . Let $E$ be the point where the line through $A$ perpendicular to $DO$ meets $\ell$ . Prove that $BE$ is tangent to the circle $\Gamma$ .

Solution

Let $F$ be the second intersection point of $DO$ and $\Gamma$ so that $DF$ is a diameter of $\Gamma$ and $BDCF$ is a cyclic quadrilateral. Then $\angle DCF = 90^\circ$ and $CF$ is the internal bisector of $\angle BCA$ as it is perpendicular to the external angle bisector $\ell$ .
Let $G$ be the intersection point of $AE$ and $DO$ . Since $BDCF$ is cyclic and $\angle AGD = \angle DBF = 90^\circ$ , we have
$\angle BCD = \angle BFD = 90^\circ - \angle BDF = \angle DAG,$
hence quadrilateral $ABCE$ is cyclic. In particular, $\angle AEB = \angle ACB$ .

$\begin{align*} \angle FBE &= \angle JEB = \angle AEB - \angle AEJ \\ &= \angle ACB - \angle ADG = 2\angle BCF - \angle BDF \\ &= \angle BCF, \end{align*}$

where we used cyclicity of $BDCF$ for $\angle BCF = \angle BDF$ . Thus $\angle FBE = \angle BCF$ , and by the Alternate Segment Theorem $BE$ is tangent to $\Gamma$ .

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