Let two circles $S_{1}$ and $S_{2}$ meet at the points $A$ and $B$. A line through $A$ meets $S_{1}$ again at $C$ and $S_{2}$ again at $D$. Let $M$, $N$, $K$ be three points on the line segments $CD$, $BC$, $BD$ respectively, with $MN$ parallel to $BD$ and $MK$ parallel to $BC$. Let $E$ and $F$ be points on those arcs $BC$ of $S_{1}$ and $BD$ of $S_{2}$ respectively that do not contain $A$. Given that $EN$ is perpendicular to $BC$ and $FK$ is perpendicular to $BD$ prove that $\angle EMF=90^{\circ}$.
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Let $G=NE\cap S_1,\ X=NE\cap MK,\ Y=KF\cap MN$. The quadrilateral $XNKY$ is ccylic because $\angle MXN=\angle MYK=\frac \pi 2$, so $\angle MNE=\angle FKM\ (*)$. If we show that the triangles $MNE,FKM$ are similar, then we would have $\angle NMK=\pi-\angle YMK=\frac \pi 2+\angle MKY=\frac \pi 2+\angle NME+\angle KMF$, which is exactly what we want.
We thus need to show that $\frac{MN}{NE}=\frac{FK}{KM}\ (**)$, which, together with $(*)$, proves the similarity of $MNE,FKM$.
$(**)\iff NE\cdot KF=BN\cdot BK\ (***)$, because $BN=MK,\ BK=MN$. Since all triangles $BCD$ are similar, no matter which line $CD$ we choose, there is a spiral similarity $\mathcal S$ cantered at $B$ which maps $D\to C$. $\mathcal S$ maps $BD\to BC$, so if $K'=\mathcal S(K)$, then $K'\in BC$ and $\frac{BK'}{K'C}=\frac{BK}{KC}=\frac{CN}{NB}$, meaning that $K'$ is the symmetric of $N$ wrt the perpendicular bisector of $BC$.
We then have $NE\cdot NG=BN\cdot NC=BN\cdot BK'$, and this proves $(***)$ because $\frac{K'F'}{KF}=\frac{BK'}{BK}$ and $K'F'=NG$, so $\frac{NG}{KF}=\frac{BK'}{BK}$.
77ant wrote:
Strangely, $ \angle EMF\neq 90^{\circ}$ in my figure.
Why does it not hold in my figure? please help me
I think that $ E$ has to be on the major arc $ CB$.
Dear Mathlinkers,
this problem appears also many time on Mathlinks.
I research a synthetic proof without calculation or transformations...
Is there one?
Sincerely
Jean-Louis
Hmm.... the problem seems to require $A$ to lie in between $C$ and $D$; I think 77ant's diagram satisfies all the conditions.
Here's a more straightforward way to establish (***) from grobber's solution...
The relation is equivalent to $\triangle{ENB}\sim\triangle{BKF}$, or $\angle{EBN}+\angle{FBK}=90^\circ$. By the Law of Sines on $\triangle{BDF}$, we have
\begin{align*}
\frac{BD}{\sin\angle{BAD}}=\frac{BF}{\sin\angle{BDF}}
&=\frac{BF}{\sin(\angle{BAD}-\angle{FBK})}\\
&=\frac{BK}{\cos\angle{FBK}\sin(\angle{BAD}-\angle{FBK})}.
\end{align*}By symmetry,
\begin{align*}
\frac{BC}{\sin\angle{BAD}}=\frac{BC}{\sin\angle{BAC}}
&=\frac{BN}{\cos\angle{EBN}\sin(\angle{BAC}-\angle{EBN})}\\
&=\frac{BN}{\cos\angle{EBN}\sin(\angle{BAD}+\angle{EBN})}.
\end{align*}Thus $NM\|BD$ gives us
\[1=\frac{CN}{CB}+\frac{BN}{BC}=\frac{NM}{BD}+\frac{BN}{BC}=\frac{BK}{BD}+\frac{BN}{BC},\]whence plugging in and simplifying yields
\[\cos(\angle{FBK}+\angle{EBN})\sin(\angle{BAD}-\angle{FBK}+\angle{EBN})=0.\]But
\[0=\sin(\angle{BAD}-\angle{FBK}+\angle{EBN})=\sin(180^\circ-\angle{BCE}-\angle{KBF})\]is impossible since $0<\angle{BCE},\angle{KBF}<90^\circ$. Thus $\angle{FBK}+\angle{EBN}=90^\circ$, as desired.
We can reconstruct the whole diagram from five points $B,C,D,M,E$. So let's fixed these points.
The existence of point $A$ ensures that $\angle CEB+\angle BFD=\angle BAD+\angle BAC=180^{\circ}$.
So point $F$ is the unique point satisfying: (1) $FK\perp BD$; (2) line $BD$ separates $F$ and $M$; (3) and $\angle CEB+\angle BFD=180^{\circ}$.
Now we assume that $F$ is the point satisfying (1), (2), and $\angle EMF=90^{\circ}$. We will prove that $\angle CEB+\angle BFD=180^{\circ}$.
We have \[\angle EMF=\angle EMN+\angle NMK+\angle KMF=\angle EMN+\angle MNC+\angle KMF=90^{\circ}-\angle MEN+\angle KMF\] so $\angle MEN=\angle KMF$. Similarly we get $\angle NME=\angle KFM$. So $\triangle MEN\sim\triangle FMK$.
Now we have $\frac{CN}{NE}=\frac{CN}{NM}\cdot\frac{NM}{NE}=\frac{MK}{KD}\cdot\frac{KF}{KM}=\frac{KD}{KF}$, so $\triangle CNE\sim\triangle FKD$. We also have $\frac{EN}{NB}=\frac{EN}{NM}\cdot\frac{NM}{NB}=\frac{MK}{KF}\cdot\frac{BK}{KM}=\frac{BK}{KF}$. Hence $\triangle EBN\sim\triangle BFK$. Therefore we conclude that \[\angle ECB=\angle CEN+\angle BEN=\angle FDK+\angle FBK=180^{\circ}-\angle BFD\] as desired.
Solution
Let $X$ be the point such that $E$ and $X$ are on opposite sides of $BC$ and $\triangle{CXB} \sim \triangle{DFB}$. First note that since $C$, $A$ and $D$ are collinear, $\angle{CXB}=\angle{DFB}=\angle{CAB}=180^\circ - \angle{CEB}$. Therefore $XBEC$ is cyclic. Let $X'$ be the point on $S_1$ such that $XX' \| BC$ and let $Y$ and $Y'$ be the projections of $X$ and $X'$, respectively, onto $BC$.
Now note that since the sides of $\triangle{MNC}$ and $\triangle{DKM}$ are parallel, it follows that
\[\frac{CN}{NB}=\frac{BK}{KD}=\frac{BY}{YC}=\frac{CY'}{BY'}\]
since $BXX'C$ is an isosceles trapezoid and $DFBK$, $CXBY$ and $BX'CY'$ are similar. The above equality implies that $Y'=N$ and therefore that $X'$ lies on $EN$. Since $BXX'C$ is an isosceles trapezoid, it follows that $BX=X'C$ and that $EX$ is the reflection of $EX'$ around the bisector of $\angle{CEB}$. Since the orthocenter of $\triangle{CEB}$, which lies on $EX'$ since $EX' \perp BC$, and the circumcenter of $\triangle{CEB}$ are isogonal conjugates, $EX$ is a diameter of $S_1$. This implies that
\[\angle{DBF}=\angle{CBX}=90^\circ - \angle{CBE}=\angle{NEB}\]
Therefore $\triangle{NEB} \sim \triangle{KBF}$ which implies that
\[\frac{EN}{KB} = \frac{BK}{KF} \quad \Rightarrow \quad \frac{EN}{NM}=\frac{MK}{KF}\]
since $MNBK$ is a parallelogram. This also implies that $\angle{MNE}=\angle{MKF}=90^\circ +\angle{CBD}$. Hence $\triangle{MNE} \sim \triangle{FKM}$ which implies that
\[\angle{EMF}=\angle{NMK}+\angle{EMN}+\angle{MFK}=\angle{CBD}+\angle{EMN}+\angle{MEN}=90^\circ\]
Hence $\angle{EMF}=90^\circ$ as desired. $\blacksquare$
Here is my proof. I hope I didn't make mistakes in using directed angles.
We'll use directed angles modulo $180$.
The key idea is that a spiral similarity $\delta$ about $B$ takes $S_2$ to $S_1$ and segment $BD$ to $BC$. This follows because $\angle BAC= \angle BAD$ implies arcs $BEC$ and $DAB$ have the same measure.
Let the image of $K$ and $F$ under $\delta$ be $K'$ and $F'$. We know $K'$ lies on $BC$, $F'$ lies on $S_1$, and $K'F' \perp BC$. Since
\[
\frac{K'C}{K'B} = \frac{DK}{KB} = \frac{AD}{AC} = \frac{NB}{NC},
\]
we also know $CK' = NB$.
We will show $\triangle MKF\sim \triangle ENM$. First, $\angle MKF = \angle MKB + \angle BKF = \angle BNM + \angle ENB = \angle ENM$. Second, because $\triangle CK'F' \sim \triangle CNE$ (because $CK' = NB$), we have
\begin{align*}
\frac{MK}{NE} &= \frac{NB}{NE} \\
&= \frac{CK'}{NE} \\
&= \frac{K'F'}{CN} \\
&= \frac{KF}{NM},
\end{align*}
so $\triangle MKF\sim \triangle ENM$. The last equality follows because $\delta$ has ratio $\frac{CB}{BD} = \frac{CN}{NM}$.
Now
\begin{align*}
\angle EMF &= \angle EMN + \angle NMK + \angle KMF \\
&= \angle EMN + \angle NMK + \angle NEM \\
&= \angle ENM + \angle NMK \\
&= 90 + \angle BNM + \angle NMK \\
&= 90,
\end{align*}
as desired.
Remark. With the above understanding it's easy to show $\angle EAF$ as well. We have
\begin{align*}
\angle EAF &= \angle EAB + \angle BAF \\
&= \angle ECB + \angle BDF \\
&= \angle ECB + \angle BCF' \\
&= \angle CF'K' + \angle K'CF' \\
&= 90,
\end{align*}
where the second-to-last equality follows because $\triangle CNE \sim \triangle F'K'C$.
If $ABC$ and $DEF$ are distinct line segments in that order such that $AB/BC=DE/DF$ and if we let points $G$ and $H$ be on the perpendiculars from $B$ to $AC$ and $E$ to $DF$ such that $\angle AGC+\angle DHF=180$, then $\triangle BGC\sim\triangle DHE$ and $\triangle AGB\sim\triangle EHF$.
Proof: Consider the point $H'$ on the perpendicular from $E$ to $DF$ such that $\triangle BGC\sim\triangle EAH'$. Then $EH'/BC=DE/BG$, implying $EH'/AB=BC/AB*DE/BG=EF/BG$ as $BC/AB=EF/DE$ from definition. Then $EH'/AB=EF/BG$, and by SAS similarity $\triangle ABG\sim\triangle H'EF$.
This implies that $\angle DH'F+\angle AGC=\angle BAG+\angle BCG+90-\angle BAG+90-\angle BCG=180$ from the similar triangles equal angles, thus $H'=H$ as there is a unique $H$. $\blacksquare$
As $MKBN$ is a parallelogram, we have $CN/CB=NM/BD=BK/BD$, implying $CN/NB=BK/BD$ with some ratio manipulating. Also, $\angle BFD+\angle BAD=180,\angle CEB+\angle CNB=180$, and $\angle CAB+\angle BAD=180$, thus $\angle CEB+\angle BFD=180$.
We know apply our lemma to find using the similar triangles $EN/BK=NB/KF$, or using the parallelogram's congruent opposite sides, $EN/MK=NM/KF$ and as $\angle ENM=90+\angle BNM=90+\angle MKB=\angle MKF$, thus by SAS similarity $\triangle NEM\sim\triangle KMF$ and
\[\angle EMF=180-\angle BNM-\angle EMN-\angle FMK=180-\angle BNM-\angle EMN-\angle MEN=180-\angle BNM-(180-90-\angle BNM)=90\]
We first prove that $\angle EBC + \angle FBD = 90^{\circ}$, and $\angle ECB + \angle FDB = 90^{\circ}$.
To do this, note that we can create a $\triangle A'B'C'$, so that there is a point $D'$ on $B'C'$ so that $\triangle A'D'B' \sim \triangle CEB$, and $\triangle A'D'C' \sim \triangle BFD$. So, we just need to prove that if $E'$ is foot of perpendicular from $D'$ to $A'B'$ and $F'$ is foot of perpendicular from $D'$ to $A'C'$, then if $E'F' || B'C'$ is given, then we have $\angle D'A'B' + \angle D'C'A' = 90^{\circ}$.
But we have $90^{\circ} = \angle D'F'A' = \angle D'F'E' + \angle E'F'A' = \angle D'C'A' + \angle D'A'B'$.
Now, this directly implies that $\angle EAF = \angle EAB + \angle BAF = \angle ECB + \angle FDB = 90^{\circ}$. So, we only need to prove that $E,M,A,F$ are concyclic.
Now, let $X = MK \cap EB$. Then $\angle XMA = \angle ACB = \angle AEB = \angle AEX$, hence $M,A,X,E$ are concyclic.
Now, using the fact that $\triangle EBN \sim \triangle BFK$, we get \[\frac{EB}{BF} = \frac{BN}{FK} = \frac{MK}{KF}\] Combining with $\angle EBF = \angle MKF$, we get $\triangle MKF \sim \triangle EBF$.
Thus, $\angle XEF = \angle XMF$, which gives $X,A,M,E,F$ concyclic, and we're done.
Let $E_1$, $F_1$ be the antipodes of $E,F$ with respect to their corresponding circles. It's not hard to see that the spiral similarity centered about $B$ sending $BC\cup S_1$ to $BD\cup S_2$ also sends $E_1$ to $F$ and $E$ to $F_1$. Therefore $A,E_1,F$ are collinear and $\angle EAF=90^\circ$. Now note that
\[
\frac{BN}{CN} = \frac{\cot\angle EBN}{\cot \angle ECN} = \frac{\sin \angle ECN}{\sin \angle EBN}\cdot \frac{\sin \angle CBE_1}{\sin \angle BCE_1} = (E,E_1;C,B)=(AE,AF;CD,AB)
\]Now $\frac{BN}{CN}=\frac{DM}{CM}$. Consider an inversion about $A$, which maps $E,F$ to points $E',F'$ on lines $B'C'$ and $B'D'$. We want to show that $E',F',$ and $M'$ are collinear. But
\[
(C',D';A,M')=\frac{DM}{CM}=(AE,AF;CD,AB)=(AE',AF';C'D',AB')=(E',F';AB' \cap E'F', C'D' \cap E'F') = (C',D';A,C'D' \cap E'F')
\]so $C'D'$ meets $E'F'$ at $M'$, as desired.
Probably the easiest G8
Let $EN$ intersect the circumcircle of $BCE$ again at $E'$, $EF'$ be a diameter of the circumcircle of $BCE$, and let $K'$ be the foot of the altitude from $F'$ to $BC$. Note that $BN=CK'$ and $E'N=F'K'$. Note that $\angle BF'C=\angle BAC=180-\angle BAD=\angle BFD$. Since $\frac{BK'}{K'C}=\frac{CN}{NB}=\frac{CM}{MD}=\frac{BK}{KD}$, $BF'CK'$ and $BFDK$ are similar. Now, note that $\frac{EN}{NC}=\frac{BN}{NE'}=\frac{CK'}{K'F'}=\frac{DK}{KF}$, and multiplying both sides by $\frac{BC}{BD}$ we have that $\frac{EN}{NM}=\frac{MK}{KF}$. Since $\angle ENM=\angle MKF=90+\angle CBD$, triangles $NEM$ and $KMF$ are similar. But since $MK\parallel BC\perp EN$ and $MN\parallel BC\perp FK$, $ME\perp MF$ as desired.
Let $EN$ intersect $S_1$ again at $E'$. Note that $\angle CE'B=\angle CAB=\angle DFB$, and $\dfrac{DK}{KB}=\dfrac{BN}{NC}$. This implies that $\triangle DFB\sim \triangle BE'C$, which you can see by fixing $B$ and $C$ on a circle and varying $A$ along an arc; clearly $\dfrac{BH}{HC}$ determines the location of $A$ where $H$ is the foot of the altitude from $A$. Then, $\angle FBD=\angle E'CB=90-\angle CBE\implies \dfrac{FK}{KB}=\dfrac{BN}{NE}\implies \dfrac{FK}{MN}=\dfrac{KM}{NE}$. Since $\angle FKM=90+\angle DKM=90+\angle MNC=\angle MNE$, so $\triangle ENM\sim \triangle MKE$. Then $\angle FME=\angle FMK+\angle KMN+\angle NME=\angle FMK+\angle DKM+\angle NFK=180-\angle FND=90$.
The problem itself is not hard. However, we must point out that the claim is true only when $A$ lies on segment $CD$. Otherwise, either $E$ or $F$ should lie on the arc containing $A$ to keep the claim true.
2001 G8 was easier
Note that $\frac{DK}{KM}=\frac{MN}{NC}$ so $\frac{DK}{KB}=\frac{BN}{NC}.$ Extend $FK$ to intersect $S_2$ at $H.$ Note that $\angle BHD=\angle BAD=\angle CEB.$ Let the transformation $S$ be from $C,B$ to $B,D$ respectively consisting of one reflection, preserving all angles and, by extension, length ratios. $N$ goes to $K$ and so the line $EN$ goes to the line $HK.$ Since $\angle CEB=\angle BHD,$ $E$ goes to $H.$
Thus, $\angle NBE=\angle HDB=\angle KFB.$ Therefore, $\angle NBE+\angle KBF=90^\circ.$ This implies that $\triangle NBE\sim\triangle KFB$ so $\frac{NE}{MK}=\frac{MK}{KF}.$ Additionally, by angle chasing $\angle MNE=\angle FKM$ so $\triangle MNE\sim\triangle FKM.$ Now, $\angle NME+\angle KMF=90^\circ-\angle MKB=\angle NMK-90^\circ,$ so we are done.
Let $FK$ meet $S_2$ again at $G,G'$ is reflection of $G$ across perpendicular bisector of $BD.$
Obviously $\measuredangle MNE=\measuredangle FKM$ and $|BN|:|NC|=|DM|:|MC|=|DK|:|KB|,$ so by spiral similarity $$BEC\stackrel{+}{\sim} BG'D\stackrel{-}{\sim} BGD\implies BNE\stackrel{+}{\sim} FKB\implies \frac{|MN|}{|FK|}=\frac{|BK|}{|FK|}=\frac{|EN|}{|BN|}=\frac{|EN|}{|KM|}\implies$$$$\implies MNE\stackrel{+}{\sim} FKM\implies \measuredangle EMF=\angle (EN,MK)=\angle (EN,BC)=90^\circ \text{ } \blacksquare$$
Consider the spiral sim $\tau$ sending $CB \mapsto BD$. Since \[\frac{BN}{NC} = \frac{DK}{KB} = \frac{DM}{MC}\]it follows that $N$ is mapped to point $K$ under $\tau$. Let $E'$ denote the map of $E$. We note that now $K, E', F$ collinear.
Consider triangle $\triangle FBD$ and note that since $E'$ lies on the $F$-altitude with \[\angle BFD + \angle BE'D = \angle BAC + \angle BEC = 180^{\circ}\]implying that $B, D, F, E'$ form an orthocentric system. Thus we have \[\angle EBN = \angle E'DK = \angle BFK\]This implies that triangles $EBN$ and $BFK$ are similar. We therefore obtain \[\frac{EN}{BN} = \frac{BK}{KF}\]Since $MNBK$ is a parallelogram, we have $MN = BK$ along with $MK = BN$. Thus we have \[\frac{EN}{NM} = \frac{EN}{BK} = \frac{BN}{KF} = \frac{MK}{KF}\]Since \[\angle ENM = 90^{\circ} + \angle CBD = \angle MKF\]we have $\triangle ENM \sim \triangle MKF$. Since $EN$ is perpendicular to $BC$ which is in turn parallel to $MK$ it follows that the corresponding sides of these triangles are at $90^{\circ}$ with each other.
In particular, $EM \perp MF$, as desired.
Actually from simple angle chase we can also deduce $\angle EAF = 90^{\circ}$, so we in fact have points $E, F, M, A$ concylic.
