, 2016

Geometry Difficulty 5.4 AIME, harder Prove it Romania

Let $ABCD$ be a cyclic quadrilateral, and let diagonals $AC$ and $BD$ intersect at $X$ . Let $C_1$ , $D_1$ and $M$ be the midpoints of segments $CX$ , $DX$ and $CD$ , respectively. Lines $AD_1$ and $BC_1$ intersect at $Y$ , and line $MY$ intersects diagonals $AC$ and $BD$ at different points $E$ and $F$ , respectively. Prove that line $XY$ is tangent to the circle through $E$ , $F$ and $X$ .

Solution

We are to prove that $\angle EXY = \angle EFX$ ; alternatively, but equivalently, $\angle AYX + \angle XAY = \angle BYF + \angle XBY$ .

Since the quadrangle $ABCD$ is cyclic, the triangles $XAD$ and $XBC$ are similar, and since $AD_1$ and $BC_1$ are corresponding medians in these triangles, it follows that $\angle XAY = \angle XAD_1 = \angle XBC_1 = \angle XBY$ .

Finally, $\angle AYX = \angle BYF$ , since $X$ and $M$ are corresponding points in the similar triangles $ABY$ and $C_1D_1Y$ : indeed, $\angle XAB = \angle XDC = \angle MC_1D_1$ , and $\angle XBA = \angle XCD = \angle MD_1C_1$ .

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.