Isosceles triangle

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

Isosceles triangle

Messaggio da stergiu » 04 set 2009, 23:05

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

Messaggio da GioacchinoA » 05 set 2009, 15:54

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

Messaggio da stergiu » 06 set 2009, 20:40

Excellent :D !!!

Babis

Rispondi