IRL ABooklet

Geometry Difficulty 6.3 National Olympiad Prove it Ireland

Two circles $\Gamma_1$ and $\Gamma_2$ intersect at two distinct points $A$ and $B$ . A line $\ell$ through $A$ meets $\Gamma_1$ and $\Gamma_2$ at points $C$ and $D$ respectively, such that $D$ lies inside $\Gamma_1$ . The line perpendicular to $\ell$ through $A$ meets $\Gamma_1$ and $\Gamma_2$ at points $E$ and $F$ respectively. The line $EC$ meets the line $FD$ (extended) at the point $X$ . Prove that if $X$ lies on the perpendicular bisector of $EF$ , then $BX$ bisects the angle $\angle CBD$ .

Solution

As $X$ lies on the perpendicular bisector of $EF$ we have $\angle XEF = \angle XFE$ ; call this angle $\alpha$ . Cyclicity of $AFBD$ implies $\angle DBA = \angle DFA = \angle XFE = \alpha$ . Cyclicity of $EABC$ gives $\angle CBA = 180^\circ - \angle CEA = 180^\circ - \angle XEF = 180^\circ - \alpha$ . Thus $\angle CBD = \angle CBA - \angle DBA = 180^\circ - 2\alpha$ .

Applying the exterior angle theorem to triangle $EFX$ we get that $\angle CXD = \angle CXF = \angle XEF + \angle XFE = 2\alpha$ . Thus $\angle CBD + \angle CXD = 180^\circ$ and it follows that the quadrilateral $DBCX$ is cyclic.

Since $\angle EAC$ and $\angle DAF$ are right triangles, we have $\angle DCX = \angle ACE = 90^\circ - \alpha$ and $\angle XDC = \angle ADF = 90^\circ - \alpha$ ( $\angle XDC$ and $\angle ADF$ being vertically opposite). Cyclicity of $DBCX$ then implies that $\angle DBX = \angle DCX = \angle XDC = \angle XBC$ , so $BX$ bisects the angle $\angle CBD$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.