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Ukrainian National Mathematical Olympiad, Third Round, Second Tour

Geometry Difficulty 4.9 AIME Prove it Ukraine

In a right triangle $ABC$ with right angle $C$ on the sides $BC$ , $AC$ and $AB$ points $D, E$ and $F$ correspondingly were chosen so that $\angle DAB = \angle CBE$ and $\angle BEC = \angle AEF$ . Prove that $DB = DF$ .
(Mykhailo Shtandenko)

Solution

Consider the point $K$ , symmetric to the point $B$ with respect to point $C$ (fig. 13). Then $\Delta BEC = \Delta KCE$ , and $\angle KEC = \angle BEC = \angle AEF$ , so points $K, E, F$ lie on the same line. Then from the statement it follows, that

Fig. 13
$\angle FAD = \angle EBC = \angle EKC = \angle FKD,$
so points $A, K, D, F$ are concyclic. Therefore
$\angle BFD = 180^\circ - \angle AFD = \angle AKD.$
As $\Delta ABC = \Delta AKC$ , and $\angle AKC = \angle ABC$ , the equality above is rewritten as $\angle BFD = \angle ABC = \angle FBD$ . From this in $\Delta FBD$ it follows that $DB = DF$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.