Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 8.1 Shortlist Prove it Baltic Way

Quadrilateral $ABCD$ is circumscribed about a circle $\omega$ . $E$ is the intersection point of $\omega$ and the diagonal $AC$ , which is nearest to $A$ . Point $F$ is diametrically opposite to point $E$ in the circle $\omega$ . The line which is tangent to $\omega$ in the point $F$ intersects lines $AB$ and $BC$ in points $A_1$ and $C_1$ , and lines $AD$ and $CD$ in points $A_2$ and $C_2$ respectively. Prove that $A_1C_1 = A_2C_2$ .

Solution

Denote by $X$ the intersection point of the lines $A_1A_2$ and $AC$ .
Prove that $X$ is a contact point of escribed circle of $\triangle AA_1A_2$ with side $A_1A_2$ . Indeed, consider a homothety with center $A$ which maps incircle $\omega$ of $\triangle AA_1A_2$ to its escribed circle. This homothety maps the line that is tangent to $\omega$ in point $E$ to the parallel line which is tangent to the escribed circle, i.e. to the line $A_1A_2$ . Therefore the point $E$ maps to the point $X$ , hence $A_1A_2$ is tangent to the escribed circle of $\triangle AA_1A_2$ in the point $X$ .

One can similarly prove that $X$ is a tangent point of the line $C_1C_2$ and incircle of $\triangle C_1CC_2$ .
From the first statement we conclude that $A_1X = FA_2$ , and from the second one that $C_1X = FC_2$ . It remains to subtract the second equality from the first one.

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