Let $ABC$ be an isosceles triangle with apex at $C.$ Let $D$ and $E$ be two points on the sides $AC$ and $BC$ such that the angle bisectors $\angle DEB$ and $\angle ADE$ meet at $F,$ which lies on segment $AB.$ Prove that $F$ is the midpoint of $AB.$
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Tags: trigonometry, geometry, angle bisector, geometry proposed
Martin N.
14.11.2010 15:52
The bisectors of angles $\sphericalangle{DEB}$ and $\sphericalangle{ADE}$ are external angle bisectors of $\triangle{CDE}$, they concur on the internal angle bisector through $C$. Hence $F$ is the intersection of $AB$ and the internal bisector of angle $\sphericalangle{ACB}$, which is the midpoint of $AB$ as $\triangle{ABC}$ is isosceles.
WakeUp
14.11.2010 16:19
Undoubtedly Martin N's solution is the easiest. Here is an alternative one:
For convenience sake, label $\angle ADF,\angle BEF,\angle BAC$ as $x,y,z$ respectively. Then $\angle DFE=180-x-y$. But also $\angle AFE=y+z$. Hence $\angle AFD=\angle AFE -\angle DFE=y+z-(180-x-y)=x+2y+z-180$. But also $\angle AFD=180-x-z$, so we get the equation $x+2y+z-180=180-x-z\implies x+y+z=180$. This means $\angle AFD=\angle FEB=y$ and $\triangle AFD$ is similar to $\triangle FEB$.
Therefore $\frac{BF}{AD}=\frac{FE}{FD}=\frac{\sin x}{\sin y}$ (the last part by the sine rule on $\triangle FED$).
Also by the sine rule on $\triangle AFD$, we have $\frac{FA}{AD}=\frac{\sin x}{\sin y}=\frac{BF}{AD}\implies BF=AF$.
sayantanchakraborty
27.01.2014 09:23
Draw perpendiculars from $F$ to $AD,BE$ and $DE$ meeting at $X,Z,Y$.Then easy to note that $\triangle DYF\cong \triangle DXF, \triangle FYE\cong \triangle FZE$,so $XF=YF=ZF$.But then $\triangle FXA\cong \triangle FZB$ so $AF=FB$ as desired. Maths is the doctor of science...... Sayantan
Onlygodcanjudgeme
04.02.2014 12:09
it is easy question. We know that F is excenter of triangle CDE,and CF is bisector of angle ACB(AC=BC)
golmakani
04.02.2014 15:37
it's not easy???????????? $F$ is excenter of $CDE$ and $CF$ is angle bisector