Mathematical and Physical Journal
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# Problem A. 495. (December 2009)

A. 495. In the acute triangle ABC we have BAC=. The point D lies in the interior of the triangle, on the bisector of BAC, and points E and F lie on the sides AB and BC, respectively, such that BDC=2, AED=90o+, and BEF=EBD. Determine the ratio BF:FC.

(5 pont)

Deadline expired on January 11, 2010.

Solution. Let G be the reflection of E through the bisector AD. Reflect point D through the lines AB and AC; denote the reflections by U and V, respectively. We have UE=DE=DG=VG.

From the deltoid AEDG we obtain and .

Moreover, and .

Since BEU=AEG and CGV=AGE, the ponts U, E, G and V are collinear.

Leet BEF=BDE=EBU=x and CGF=CDG=GCV=z. From the quadrilateral ABDC and the triangle ABC, we have x+y=360o--(360o-2)= and

UBC+BCV=(ABC+x)+(ACB+y)=180o.

Therefore, the lines BU and CV are parallel. Due to EBU=VEF, also the line EF is parallel with them.

Hence,

### Statistics:

 5 students sent a solution. 5 points: Bodor Bertalan, Éles András, Nagy 235 János, Nagy 648 Donát. 2 points: 1 student.

Problems in Mathematics of KöMaL, December 2009