## Isosceles triangle

Rette, triangoli, cerchi, poliedri, ...
stergiu
Messaggi: 8
Iscritto il: 01 set 2009, 18:41

### Isosceles triangle

Well, a nice problem for beginers with olympiads is the following:

Problem

The incircle of a triangle ABC touches BC at D. Let E be the projection of B on the bisector of angle A. If M is the midpoint of side BC, prove that the triangle MDE is isosceles.

Enjoy it - Babis

GioacchinoA
Messaggi: 137
Iscritto il: 13 feb 2009, 15:44
Località: Bari
Let $F,G$ be the points where the incircle touches sides $AB$ and $AC$ and let $N$ be the midpoint of $AB$.

Since $ABE$ si right angled triangle, $EN = \dfrac{AB}{2}$.
Moreover $\angle NEA = \angle NAE = \angle EAC$, thus $NE$ is parallel to $AC$. This obviously implies that $N,E,M$ are collinear.
So $ME = |MN - NE| = |\dfrac{AC-AB}{2}|$.

Let us consider $DM$ wich obviously equals $|BM-BD| = |\dfrac{BC}{2} - BD|$.
Now, we have to calculate $BD$.
Notice that $BD=BF,AF=AG,CG=CD;BD+CD=BC;$
$BF+AF=AB;AG+CG=AC$. From this easily follows $BD = \dfrac{BC+AB-AC}{2}$, thus $DM = |BM-BD| = |\dfrac{BC}{2} - \dfrac{BC+AB-AC}{2}| = |\dfrac{AC-AB}{2}|$.

Since $ME = DM$ the triangle $MDE$ is isosceles(with base $DE$)

stergiu
Messaggi: 8
Iscritto il: 01 set 2009, 18:41
Excellent !!!

Babis