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