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#1 h Apr 14, 2006, 8:47 PM • 3 Y

Y by mathematicsy, Adventure10, Mango247

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 $\sphericalangle{ABC}$ . 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 $\sphericalangle{EMD}=\sphericalangle{DMF}$ .

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#2 h Apr 15, 2006, 1:07 AM • 2 Y

Y by Adventure10 and 1 other user

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.

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#3 h Apr 26, 2006, 10:43 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

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

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#4 h Feb 10, 2010, 5:33 PM • 3 Y

Y by mathematicsy, Adventure10, Mango247

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 $\Longrightarrow$ $\angle EMD = \angle DMF.$

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#5 h Jul 30, 2010, 1:31 PM • 2 Y

Y by Adventure10, Mango247

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

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#6 h Nov 19, 2011, 5:53 AM • 3 Y

Y by wandarboux, Adventure10, Mango247

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

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#7 h Apr 19, 2015, 1:52 PM • 3 Y

Y by DominicanAOPSer, Adventure10, Mango247

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 $\angle DMF=\angle DCF=\angle DAE=\angle EPD=\angle EMD$ .

This post has been edited 1 time. Last edited by JoseT17, Apr 19, 2015, 1:52 PM

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#8 h Dec 17, 2017, 12:33 AM • 3 Y

Y by helol, Adventure10, Mango247

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.

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#9 h Apr 12, 2021, 6:39 PM

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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.

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#10 h Sep 28, 2021, 8:32 PM

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#11 h Oct 9, 2021, 11:38 PM

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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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This post has been edited 5 times. Last edited by parmenides51, Oct 10, 2021, 12:14 AM

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