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Ukrainian National Mathematical Olympiad, Third Round, Second Tour

Geometry Difficulty 5.3 AIME, harder Prove it Ukraine

Let $ABC$ be a right triangle with hypothenuse $BC$ and altitude $AD$ . Let's denote the midpoints of $AD$ and $AC$ by $E$ and $F$ correspondingly. Let point $M$ be the circumcenter of $\triangle BEF$ . Prove that $AC||BM$ .

Solution

As $\triangle ADB \sim \triangle CAB$ , we get $\frac{AD}{AB} = \frac{CA}{CB}$ . As $AE = \frac{1}{2}AD$ , $CF = \frac{1}{2}CA$ , we get $\frac{AF}{AB} = \frac{CF}{CB}$ . From this similarity we get that (fig. 16):
$\angle BAE = \angle BAD = \angle BCA = \angle BCF$
Then by angle and the ratio of the sides we get that $\triangle AEB \sim \triangle CFB \Rightarrow \angle ABE = \angle CBF$ .
As $EF$ is the midline of $\triangle CAD$ , we also get $EF||BC \Rightarrow \angle BFE = \angle CBF$ . Next, we get the following equalities of angles: $\angle BFE = \frac{1}{2}\angle BME = 90^\circ - \angle EBM \Rightarrow \angle ABE = \angle CBF = 90^\circ - \angle EBM$ . Therefore $\angle ABM = \angle ABE + \angle EBM = 90^\circ, AC||BM$ .

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