# 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*=90^{o}+, 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*=360^{o}--(360^{o}-2)= and

*UBC*+*BCV*=(*ABC*+*x*)+(*ACB*+*y*)=180^{o}.

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