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Geometry Difficulty 6.7 National Olympiad Prove it Slovenia

Let $\mathcal{K}$ be a circle with centre $O$ , and let $\mathcal{K}'$ be a circle that goes through the point $O$ with a radius that is greater than twice the radius of the circle $\mathcal{K}$ . A common tangent of the circles $\mathcal{K}$ and $\mathcal{K}'$ touches the circle $\mathcal{K}$ at point $A$ , and it touches the circle $\mathcal{K}'$ at point $B$ . Let $C$ denote the mirror image of the point $B$ with respect to point $A$ . The line $AO$ intersects the circle $\mathcal{K}'$ at points $O$ and $D$ , and line $CD$ intersects circle $\mathcal{K}'$ at points $D$ and $E$ . Prove that line $BE$ is a tangent of the circle $\mathcal{K}$ .

Solution

Let the tangent from $B$ to the circle $K$ (distinct from the tangent $AB$ ) touch the circle $K$ at $A'$ . We show that the points $B$ , $A'$ and $E$ are collinear.

We have $|AB| = |A'B|$ and $|AO| = |A'O|$ , so the triangles $ABO$ and $A'BO$ have three equal sides and are therefore congruent. This implies $\angle OBA' = \angle ABO$ . By the Tangent-Chord Theorem in the circle $K'$ we have $\angle ABO = \angle BDO$ . The line $AB$ is tangent to $K$ , so $\angle OAB = 90^\circ$ . We have $|AB| = |AC|$ , so the triangles $ABD$ and $ACD$ match in two sides and the angle between them. We conclude that they are congruent and

$\angle BDO = \angle BDA = \angle ADC = \angle ODE$ . The angles over the same chord of $K'$ are equal, so $\angle ODE = \angle OBE$ .

We have shown that $\angle OBA' = \angle OBE$ , which implies that the points $B$ , $A'$ and $E$ are colinear and the line $BE$ is tangent to $K$ .

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