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
Isosceles triangle
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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 $)
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 $)
!!!