Let $F_1$ be the reflection of $F$ in $BC$. The conclusion we want to reach is equivalent to the fact that $F_1,M,E$ are collinear, so this will be the aim. The triangles $BF_1C,BMD,BEA$ have right angles in $F_1,M,E$ respectively, and the angles $\angle F_1BC,\angle MBD,\angle EBA$ are all equal to half the angle $\angle ABC$. In particular, they are equal, so the three mentioned triangles are similar. This means that $F_1,M,E$ are the images of $C,D,A$ respectively through a spiral similarity centered at $B$. Since $C,D,A$ are collinear, so are $F_1,M,E$, exactly what we wanted to show.

A different solution: We will use directed angles modulo 180°. Consider not only the orthogonal projection M of the point D on the line BC, but also the orthogonal projection N of the point D on the line AB. Since the line BD is the angle bisector of the angle ABC, these orthogonal projections M and N are symmetric to each other with respect to this angle bisector BD, and thus we have < EMD = - < END. Since < AED = 90° and < AND = 90°, the points E and N lie on the circle with diameter AD, so that < END = < EAD. Since the lines AE and CF are both perpendicular to the line BD, they are parallel to each other, and thus < DAE = < DCF. Finally, since < CFD = 90° and < CMD = 90°, the points F and M lie on the circle with diameter CD, and therefore < DCF = < DMF. Hence, < EMD = - < END = - < EAD = < DAE = < DCF = < DMF, and we are done. darij

carlosbr wrote: In a triangle $\triangle{ABC}$ with all its sides of diferent length, $D$ is on the side $AC,$ such that $BD$ is the angle bisector of $\sphericalangle{ABC}.$ Let $E$ and $F,$ respectively, be the feet of the perpendicular drawn from $A$ and $C$ to the line $BD$ and let $M$ be the point on $BC$ such that $DM$ is perpendicular to $BC.$ Show that $\sphericalangle{EMD}=\sphericalangle{DMF}.$ Let $ D'$ be the foot of the external bisector of $ \angle ABC.$ Since the division $ (A,C,D,D')$ is harmonic, then its orthogonal projection onto the internal bisector of $ \angle ABC$ is harmonic as well $ \Longrightarrow$ $ (E,F,D,B) = - 1.$ From $ DM \perp BC$ we conclude that rays $ MD$ and $ MB$ bisect $ \angle EMF$ internally and externally $ \Longrightarrow$ $ \angle EMD = \angle DMF.$

From $\triangle ABE \sim \triangle CBF$ and $\triangle ADE \sim \triangle CDF$ we get $(E, F, D, B) = -1$, further see above. Best regards, sunken rock

let N be the symmetric point of M in respect of BD,then $\angle EMD=\angle END=\angle EAD=\angle DCF=\angle DMF$.

Let $P =BC\cap AE$. We can easily see that triangle $DEP$ is congruent to triangle $DEA$ and that $DEPM$ is a cyclic cuadrilateral hence $\angle DAE=\angle EPD=\angle EMD$. Now, since $FDMC$ is a cyclic cuadrilateral, which can be proven easily, $\angle DMF=\angle DCF$ and since $CF\parallel EA$, $\angle DCF=\angle DAE$ hence $\angle DMF=\angle DCF=\angle DAE=\angle EPD=\angle EMD$.

Since $\angle DFC =\angle DMC=90, DMCF$ is a cyclic quadrilateral and thus $\angle DCF=\angle DMF$. Also, $AE||FC$ so $\angle DCF=\angle DAE$. Let $M'$ be the foot of the perpendicular to $AB$ from $D$, note that $M'EDA$ is cyclic because $\angle DM'A= \angle AED$ and thus $\angle DAE= \angle DM'E$. Finally, since $BD$ is an angle bisector, $\Delta DEM' \equiv \Delta DEM$ and therefore $\angle DM'E= \angle EMD$ completing the proof.

Attachments:

Let $G\in AE\cap BC$ and let $P_{\infty}$ be the point at infinity of lines $AE$ and $FC$. We have that $\Delta GBA$ is isosceles, since $\overline{BE}$ is both an altitude and an angle bisector. Then, $E$ is the midpoint of $AG$. Hence, $-1=(A,G;E,P_{\infty})\stackrel{C}{=}(D,B;E,F)$. Since $\angle DMB=90^\circ$, we have that $MD$ bisects angle $\angle FME$, as desired.

Sketch

In a triangle $\triangle{ABC}$ with all its sides of different length, $D$ is on the side $AC$, such that $BD$ is the angle bisector of $\angle{ABC}$. Let $E$ and $F$, respectively, be the feet of the perpendicular drawn from $A$ and $C$ to the line $BD$ . Let $M$ any point the point on $BC$ . Prove that $DM \perp BC$ iff $\angle{EMD}=\angle{DMF}$.

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