## #24

Problem: Let $BD$ be the angle bisector of $\angle ABC$ in scalene $\triangle ABC$. The points $E, F$ are on $BD$ such that $AE\perp BD, CF\perp BD$. The foot of the perpendicular from $D$ to $BC$ is $M$. Prove that $\angle EMD=\angle DMF$.
Solution: Note that $\triangle AED\sim\triangle CFD$, so $\frac{ED}{DF}=\frac{AD}{DC}$. Also, $\triangle ABE\sim\triangle CBF$. Thus $\frac{AB}{BC}=\frac{BE}{BF}$. Finally $\frac{AD}{DC}=\frac{AB}{BC}$ from Angle Bisector Theorem. So $\frac{ED}{DF}=\frac{AD}{DC}=\frac{AB}{BC}=\frac{BE}{BF}$, and it follows that $\frac{BE}{ED}=\frac{BF}{FD}$, so $(B, D; E, F)=-1$. Since $\angle BMD=90^\circ$, this implies that $MD$ bisects $\angle EMF$ as desired.

Source: OIM 2002 #4

## #23

Problem: In acute triangle $ABC$ angle $B$ is greater than$C$. Let $M$ is midpoint of $BC$. $D$ and $E$ are the feet of the altitude from $C$ and $B$ respectively. $K$ and $L$ are midpoint of $ME$ and $MD$ respectively. If $KL$ intersect the line through $A$ parallel to $BC$ in $T$, prove that $TA=TM$

Solution: By some angle chasing, we note that $\angle MED=\angle MDE=A$, so that $MD, ME$ are tangents to the circle $ADE$. Also $TA$ is tangent to this circle by the same angle chase. Let $\omega_1, \omega_2$ be circle $ADE$ and the degenerate circle $M$ respectively. Note that $KL$ is the radical axis of $\omega_1, \omega_2$, so that the power of $T$ wrt $\omega_1$ is equal to that of $\omega_2$. But the first if $TA$, and the second is $TM$, so $TA=TM$ as desired.

Source: Iran TST 2011

## #22

Problem: Let $D$ be a point on the $A-$median of acute $\triangle ABC$ such that $\angle BDC=90^\circ$. Let the foot of the altitude from $D$ to $BC$ be $H$, and let the feet of the altitudes from $H$ to $BD, H$ to $CD$ hit $AB, AC$ at $X, Y$ respectively. Prove that $BC$ is tangent to the circumcircle of $XHY$.

Solution: Let $CD\cup AB=K, BD\cup AC=L$ and let $M$ be the midpoint of $BC$. Let $BD\cup HX=P, CD\cup HY=Q$ Note that by Ceva’s on $A$ wrt $\triangle BDC$, we have:

$(\frac{CM}{MD})(\frac{DN}{NB})(\frac{BM}{MC})=1\implies \frac{BD}{BN}=\frac{CD}{CM}\implies \frac{HP}{HX}=\frac{HQ}{HY}\implies \triangle XHY\sim\triangle PHQ$

Thus $\angle XHB=90^\circ-\angle PHD=\angle XYH$, and the result follows.

Source: ISL, unknown date

## #21

Problem: Let scalene $\triangle ABC$ have an intouch triangle $DEF$, and let $EF$ intersect the perpendicular to $AD$ at $D$ at a point $X$. Prove that $BC$ bisects $AX$.

Solution: Let $AX\cap BC=K$Let $I$ be the incenter of $\triangle ABC$, and let $M=AI\cap EF$ be the midpoint of $EF$. Now we have $\angle AMX=\angle ADX=\frac{\pi}{2}$, so $AMDX$ is cyclic. Now also $\triangle IME\sim\triangle IEA$ so $ID^2=IE^2=(IM)(IA)$, and thus $\triangle IDM\sim\triangle IAD\implies \angle IDA=\angle IMD$. Now using these facts:

$\angle KXD=\pi-\angle AMD=\angle IMD=\angle IDA=\frac{\pi}{2}-\angle ADK\angle KDX$.

Thus $KD=KX$, and so $DK$ is a median to the hypotenuse in right triangle $ADX$. Thus $AK=KD$, as desired.

Source: Jean-Louis Ayme

## #20

Problem: Let the $A-$symmedian and $B-$symmedian of acute $\triangle ABC$ meet $BC, AC$ at $P, R$ respectively. Let $Q, S$ be the midpoints of $AP, BR$ respectively. Prove that $\angle BAS=\angle ABM$.

Solution: Note that $\triangle ABC$ is acute, so $0<\angle BAS, \angle ABQ<\frac{\pi}{2}$ so it suffices to show that $\sin BAS=\sin ABQ$ Now:

$\sin BAS=\frac{2[ABS]}{AB*BS}=\frac{2[ABR]}{c\sqrt{c^2+AR^2+2(c)(AR)\cos A}}=(2\frac{[ABC]}{c})(\frac{\frac{AR}{AC}}{\sqrt{c^2+AR^2+2c(AR)(\frac{b^2+c^2-a^2}{2bc})}}$

By the Law of Cosines. By Steiner’s we have $AR=\frac{bc^2}{a^2+c^2}$, so this is just:

$\sin BAS(\frac{c}{2[ABC]})=\frac{\frac{c^2}{a^2+c^2}}{\sqrt{c^2+\frac{b^2c^4}{(a^2+c^2)^2}+\frac{b^2+c^2-a^2}{a^2+c^2}}}=\frac{c}{\sqrt{(a^2+c^2)^2+b^2c^2+(a^2+c^2)(b^2+c^2-a^2)}}=\frac{c}{\sqrt{2c^4+2c^2a^2+2c^2b^2+a^2b^2}}$

Thus $\sin BAS=\frac{[ABC]}{\sqrt{2c^2(a^2+b^2+c^2)+a^2b^2}}$, which is symmetric upon switching $a, b$. Thus by symmetry $\sin BAS=\sin ABQ$, so $\angle BAS=\angle ABQ$ as desired.

Source: ?

## #19

Problem: Let $\triangle ABC$ satisfy $AB, and let $M, N$ be the mispoint of $AC$ and the midpoint of arc $ABC$ of the circumcircle $\omega$ of $\triangle ABC$, respectively. Prove that $\angle IMB=\angle INB$.

Solution: Let $X$ be the midpoint of arc $AC$. Note that $B, I, E$ are collinear, as are $E, M, N$. Therefore $\angle EIA=\pi-\angle BIA=\angle IAB+\angle IBA=\angle IAC+\angle EBC=\angle IAC+\angle EAC=\angle EAI$. Therefore $EA=EI$. Now by this and POP we have:

$(EM)(EN)=EM^2+(EM)(MN)=(AM)(MC)+EM^2=AM^2+EM^2=AE^2=IE^2$.

Thus $\frac{EM}{EI}=\frac{EI}{EN}$, so by $SAS$ similarity, $\triangle EMI\sim\triangle EIN$. Finally:

$\angle IMB=\frac{\pi}{2}-\angle IMN=\angle IME-\frac{\pi}{2}=\angle NIE-\frac{\pi}{2}=\frac{\pi}{2}-\angle BIN=\angle INB$

As desired.

Source: All-Russian MO 2005, Round 4

## #18

Problem: Let $D, E, F$ be points on segments $BC, CA, AB$ respectively in $\triangle ABC$, and let $DE, DF$ intersect the line through $A$ parallel to $BC$ at $X, Y$ respectively. Let $EY, FX$ intersect at $G$, and let $AG, BC$ intersect at $H$. Prove that $BD=HC$.

Solution: Note that by the Law of Sines, we know that $FX\sin AFG=FX\sin AFX=AX\sin FAX$. Similarly, $FX\sin EFG=FX\sin EFX=EX\sin FEX$. Therefore $\frac{\sin AFG}{\sin EFG}=(\frac{AX}{EX})(\frac{\sin FAX}{\sin FEX})$. Similarly, $\frac{\sin FEG}{\sin AEG}=(\frac{FY}{AY})(\frac{\sin EFY}{\sin EAY})$. Now, the extended angle bisector theorem:

$\frac{BH}{HC}=(\frac{AB}{AC})(\frac{\sin BAH}{\sin CAH})=(\frac{AB}{AC})(\frac{\sin FAG}{\sin EAG})$.

By Trig Ceva’s and the above lemma, this becomes:

$\frac{BH}{HC}=(\frac{AB}{AC})(\frac{\sin AFG}{\sin EFG})(\frac{\sin FEG}{\sin AEG})=(\frac{AB}{AC})(\frac{AX}{EX})(\frac{FY}{AY})(\frac{\sin FAX}{\sin EAY})(\frac{\sin EFY}{\sin FEX})$

Now, note that $\triangle FBD\sim\triangle FAY, \triangle ECD\sim\triangle EAX$. Thus $\frac{AX}{EX}=\frac{CD}{ED}, \frac{FY}{AY}=\frac{FD}{BD}$. Now we also have $\angle FAX=\pi-\angle ABC, \angle EAY=\pi-\angle BCA$, so $\frac{\sin FAX}{\sin EAY}=\frac{\sin B}{\sin C}=\frac{AC}{AB}$. Finally, $\frac{\sin EFY}{\sin FEX}=\frac{\sin EFD}{\sin FED}=\frac{ED}{FD}$. Thus we have:

$\frac{BH}{HC}=(\frac{AB}{AC})(\frac{CD}{DE})(\frac{FD}{BD})(\frac{AC}{AB})(\frac{ED}{FD})=\frac{CD}{BD}$

And the result follows.

Source: Own.