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


Problem: In acute triangle ABC angle B is greater thanC. 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


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


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=KLet 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


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: ?


Problem: Let \triangle ABC satisfy AB<BC, 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:


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


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:


And the result follows.

Source: Own.