Two circles intersect at points $A\neq B$. A line passing through $A{}$ intersects the circles again at $C$ and $D$. Let $E$ and $F$ be the midpoints of the arcs $\overarc{BC}$ and $\overarc{BD}$ which do not contain $A{}$ and let $M$ be the midpoint of the segment $CD$. Prove that $ME$ and $MF$ are perpendicular.
Problem
Source: Romania EGMO TST 2021 Day 1 P2
Tags: geometry, romania
Davsch
24.11.2022 02:39
It suffices to show that $A,M,E,F$ are concyclic, since then \[\angle EMF=\angle EAF=\angle EAB+\angle BAF=\frac 12\angle CAB+\frac 12\angle BAD=90^\circ.\]Invert around a circle centered at $A$ with arbitrary radius. The original problem transforms to the following: Let $BCD$ be a triangle and $A$ a point on $CD$. Choose $E,F$ on $BC,BD$ such that $AE,AF$ bisect $\angle CAB,\angle BAD$. Let $M$ be on $CD$ such that $(C,D;A,M)=-1$. Then $M$ lies on $EF$. To show this, it suffices to show that $BA,DE,CF$ concur. However, by angle-bisector theorem, $\frac{CE}{EB}=\frac{AC}{BA}$ and $\frac{BF}{FD}=\frac{AB}{AD}$. Multiplying these two together and applying Ceva yields the desired result.
cursed_tangent1434
09.10.2024 09:20
Pretty much the same solution as above. This was a pretty cool problem. We first need to show the following key claim. Claim : Points $A,F,E$ and $M$ are concyclic. Proof : Denote by $M'$ the second intersection of $(AEF)$ and $\overline{CAD}$. We wish to show $(CD;M'P_{\infty})=-1$. Now, consider an inversion centered at $A$. Then, it suffices to show the following. Inverted Problem wrote: Consider $\triangle BCD$ with point $A$ on side $CD$. Let $E$ and $F$ be the intersection of the internal $\angle BAC$ and $\angle BAD$ bisectors with sides $BC$ and $BD$ respectively. Let $M = \overline{EF} \cap \overline{CD}$. Show that $(CD;AM)=-1$. We solve this as follows. Note that by the Angle Bisector Theorem we have, \[\frac{BF}{FD} \cdot \frac{DA}{AC} \cdot \frac{CE}{EB} = \frac{AB}{AD} \cdot \frac{DA}{AC} \cdot \frac{CA}{AB}=1\]which by the converse of Ceva's Theorem implies that cevians $\overline{CF}$ , $\overline{DE}$ and $\overline{AB}$ concur. Thus, by the Ceva/Menalaus picture we have $(CD;AM)=-1$ as we set out to show. Now, due to the well known fact that inversion preserves the cross ratio, we must have that $(CD;AM')=-1$ as well, and thus $M'\equiv M$. This means $M$ indeed lies on $(AEF)$ and the claim is proved. To finish off note that since in $\angle BAC$, $\overline{AE}$ and $\overline{AF}$ are respectively the internal and external angle bisectors, it follows that $\angle EAF = \frac{\pi}{2}$. Thus, $\angle EMF = \frac{pi}{2}$ as well, implying that $EM \perp FM$ as desired.