angle bisection

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Source: Italy TST 2000

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#1 h Sep 27, 2008, 4:04 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an isosceles right triangle and $M$ be the midpoint of its hypotenuse $AB$ . Points $D$ and $E$ are taken on the legs $AC$ and $BC$ respectively such that $AD=2DC$ and $BE=2EC$ . Lines $AE$ and $DM$ intersect at $F$ . Show that $FC$ bisects the $\angle DFE$ .

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#2 h Sep 27, 2008, 7:52 PM • 2 Y

Y by Adventure10, Mango247

$N: AE \cap CM$ , $G: CF \cap BA$ . For Menelaus on CMB with trasversal NE we obtaid $\overline{CN}=\overline{NM}$ and for Ceva on ACM and point F we obtain $\overline{AG}=2 \overline{GM}$ and so $\overline{GB}=\overline{GA}$ . For Menelaus on AEB and trasversal GF we have $\frac{\overline{AF}}{\overline{FE}}= \frac{3}{2}$ and so $\frac{\overline{AF}}{\overline{AE}}= \frac{3}{5}$ . Let $CA=CB=x$ so $\overline{AD}=\frac{2}{3} x$ and $\overline{AE}= \frac{\sqrt{10}}{3} x$ so $\frac{\overline{AE}}{\overline{AD}} = \frac{\sqrt{10}}{2}= \frac{\overline{AC}}{\overline{AF}}$ so AFD and ACE are similar and $AE \perp DM$ , then DCEF is cyclic with DC = CE so C stay on the angle bisector of $\angle DFE$

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#3 h Oct 6, 2008, 9:18 AM • 1 Y

Y by Adventure10

See that AM^2-AD^2=ME^2-DE^2, therefore AE and DM are perpendicular, CDFE cyclic with CD=CE, so CF is angle bisector of DFE.

Best regards,
sunken rock

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#4 h Jan 15, 2014, 12:46 AM • 1 Y

Y by Adventure10

Let $C(0,0)$ , $B(0,a)$ and $A(a,0)$ .
Easily, $D(\dfrac{a}{3},0)$ , $E(0,\dfrac{a}{3})$ , $M(\dfrac{a}{2},\dfrac{a}{2})$
Let $F(p,q)$ . By solving a system,
$\begin{vmatrix} p & q &1 \\ \dfrac{a}{2} & \dfrac{a}{2} & 1 \\ \dfrac{a}{3} & 0 & 1 \end{vmatrix}=0$

$\begin{vmatrix} p & q &1 \\ 0 & \dfrac{a}{3} & 1 \\ a & 0 & 1 \end{vmatrix}=0$

we easily get $p=\dfrac{2a}{5}$ and $q=\dfrac{a}{5}.$
$k_{EF}=-\dfrac{1}{3}$
$k_{CF}=\dfrac{1}{2}$
$k_{DF}=3$
$tg(\angle EFC)=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{1-\dfrac{1}{2}\dfrac{1}{3}}=1$
$tg(\angle CFD)=\dfrac{3-\dfrac{1}{2}}{1+3\cdot\dfrac{1}{2}}=1$
$\implies \angle EFC=\angle CFD$ and we are done.

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#5 h Jan 15, 2014, 1:21 AM • 2 Y

Y by Adventure10, Mango247

Wait isnt this completely obvious by symmetry or am I messing up my diagram? Isosceles right triangle?

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#6 h Jan 15, 2014, 1:22 AM • 2 Y

Y by Adventure10, Mango247

Oh wait nvm misread problem.

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#7 h Jan 15, 2014, 9:20 AM • 3 Y

Y by pretzel.p, Adventure10, Mango247

Because reviving the thread:
Take $N$ symmetrical of $B$ about $C; D$ is the centroid of $\triangle ABN\implies N-D-M$ collinear, so $\angle ADM=\angle CDN$ , but $\angle CDN=\angle BDC$ by symmetry; also by symmetry $\angle AEC=\angle CDB$ , so $\angle ADM=\angle FEC$ and $CDFE$ is cyclic; with $CD=CE$ we are done.

Best regards,
sunken rock

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