Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.7 National Olympiad Prove it Italy

Problem:

Let $\Gamma$ be the excircle of triangle $ABC$ opposite vertex $A$ , that is, the circle tangent to $BC$ and to the extensions of sides $AB$ and $AC$ on the side of $B$ and of $C$ . Let $D$ be the center of $\Gamma$ and let $E$ and $F$ be, respectively, the points of tangency of $\Gamma$ with the extensions of sides $AB$ and $AC$ . Let $J$ be the intersection of segments $BD$ and $EF$ .
Prove that the angle $\angle CJB$ is right.

Solution

Solution:

The lines $BD$ and $CD$ are the bisectors of angles $\widehat{EBC}$ and $\widehat{BCF}$ ; letting $\alpha, \beta, \gamma$ be the interior angles of triangle $ABC$ (respectively at $A, B, C$ ), we thus know that $\widehat{BCD} = \frac{180^{\circ} - \gamma}{2}$ and $\widehat{CBD} = \frac{180^{\circ} - \beta}{2}$ . It follows (by subtraction) that $\widehat{BDC} = \frac{\beta + \gamma}{2}$ .

Triangle $AEF$ is isosceles with base $EF$ , hence $\widehat{AEF} = \widehat{EFA} = \frac{180^{\circ} - \alpha}{2} = \frac{\beta + \gamma}{2} = \widehat{BDC}$ , and therefore the quadrilateral $D J C F$ is inscribable in a circle.

On the other hand, since the angle $\widehat{DFA}$ is right (as $AF$ is tangent to the circle $\Gamma$ ), this implies that the angle $\widehat{CJD}$ , supplementary to $\widehat{DFC}$ , is also right. This obviously also implies that the angle $\widehat{CJB}$ is right.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.