Fermat -Euler wrote: $C$ and $D$ are points on a semicircle. The tangent at $C$ meets the extended diameter of the semicircle at $B$, and the tangent at $D$ meets it at $A$, so that $A$ and $B$ are on opposite sides of the center. The lines $AC$ and $BD$ meet at $E$. $F$ is the foot of the perpendicular from $E$ to $AB$. Show that $EF$ bisects angle $CFD$ Maybe it's another midnight hallucination, but I have some strange feeling of seeing the following solution on ML: Let the tangents to the semicircle at the points C and D meet each other at a point G. Also, let U be the center of the semicircle. Then, the points A and B, both lying on the diameter of this semicircle, must be collinear with its center U. Since the semicircle touches the sides AG and BG of the triangle ABG, its center U must lie on the angle bisector of the angle between these sides, i. e. of the angle AGB. In other words, the line GU is the angle bisector of the angle AGB. Since the line BG is the tangent to the semicircle at the point C, while U is the center of this semicircle, we have $BG\perp UC$; thus, the point C is the foot of the perpendicular from the point U to the line BG. Similarly, the point D is the foot of the perpendicular from the point U to the line AG. Now, we can apply the result of http://www.mathlinks.ro/Forum/viewtopic.php?t=5460 to the triangle ABG with the point U on its side AB, and obtain that the line GE is an altitude of the triangle ABG if and only if the line GU is an interior angle bisector of the triangle ABG. But the line GU indeed is an interior angle bisector, as we saw above; thus, the line GE is an altitude of triangle ABG. In other words, the point E lies on the altitude of triangle ABG from the vertex G to the side AB. Hence, the point F, being the foot of the perpendicular from the point E to the line AB, must be the foot of this altitude. Now we can describe our situation as follows: The point F is the foot of the altitude of the triangle ABG from the vertex G, and the point E is a point on this altitude GF. The lines AE and BE intersect the lines BG and AG at the points C and D, respectively. Now, according to the Blanchet theorem ( http://www.mathlinks.ro/Forum/viewtopic.php?t=20809 post #5 problem (a)), we have < DFG = - < CFG (with directed angles modulo 180°); in other words, the line EF bisects the angle CFD. And the problem is solved. An alternative approach would be possible by reflecting the whole figure in the line AB and applying some degenerate cases of Brianchon. Darij

it has a solution whit using fermat point just try it.

mohamadhosein wrote: it has a solution whit using fermat point just try it. That would be very strange. darij

$\blacksquare \ 1^{\circ}.\ Z\in BC\cap AD;\ U\in AB\cap CD;\ P\in AB,\ ZP\perp AB;$ $T\in CD\cap ZP\Longrightarrow$ quadrilaterals $PCZO,\ PZDO$ are cyclic $\Longrightarrow$ $\widehat {BPC}\equiv\widehat {BZO}\equiv\widehat {AZO}\equiv \widehat {APD}\Longrightarrow\widehat {BPC}\equiv \widehat {APD}\Longrightarrow$ $(U,C,T,D)-h.d.\Longrightarrow (U,B,P,A)-h.d.$ $\blacksquare \ 2^{\circ}.\ R\in ZE\cap AB,\ U\in AB\cap CD\Longrightarrow (U,B,R,A)-h.d.$ $\blacksquare \ 3^{\circ}.\ (U,B,P,A)-h.d.,\ (U,B,R,A)-h.d.\Longrightarrow$ $P\equiv R\equiv F\ (E\in ZP)\Longrightarrow\widehat {BFC}\equiv \widehat {AFD}\Longrightarrow\overline {\underline {\left| \ \widehat {EFC}\equiv \widehat {EFD}\ \right| }}$ Remark. I note $h.d.$ - harmonical division.

Extend $AD$ and $BC$ to meet at $P$,and let $Q$ be the foot of the perpendicular from $P$ to $AB$. Denote $O$ be the center of semicircle.We have: $\begin{cases}\triangle PAQ\sim\triangle OAD\\ \triangle PBQ\sim\triangle OBC\end{cases}\Longrightarrow\frac{AQ}{AD}=\frac{PQ}{OD}=\frac{PQ}{OC}=\Longrightarrow\frac{AQ}{QB}\cdot\frac{BC}{CP}\cdot\frac{PD}{DA}=1$ $\Longrightarrow AC,\,BD,\,BQ\,\text{are\,concurrent}\Longrightarrow Q\equiv F$ Finally, since the points $O,\,C,\,P,\,D,\,F$ are concyclic, we have: $\angle{DFP}=\angle{DOP}=\angle{POC}=\angle{PFC}$

Let $E$ and $F$ be on the semicircle so that $B,F,E,A$ are collinear, in that order. Let the tangents to the semicircle at $C$ and $D$ intersect at $P$, and let $Q$ be the intersection of $FC$ and $PE$ (possibly the point at infinity.) By Pascal's theorem on $DDECCF$, we get that $P$, $Q$, and $E$ are collinear. $EC \perp QF$ and $FD \perp QE$, so $E$ is the orthocenter of $\triangle QFE$. Hence, $Q$, $E$, and $F$ are collinear, so $P$, $E$, and $F$ are collinear. Since $PF$ is an altitude, Blanchet's theorem tells us that $\angle CFE = \angle DFE$, so our proof is complete.

My previous solution is completely incorrect... sorry about that. Lemma: In cyclic quadrilateral $ABCD$, let $AD \cap BC = P$, $AB \cap CD = Q$, $AA \cap BB = R$, $CC \cap DD = S$, and $AC \cap BD = T$. Then $P$, $R$, and $S$ each lie on the polar of $Q$. Proof: By Pascal's theorem on $AACBBD$, we find that $P$, $R$, and $S$ are collinear. By Pascal's theorem on $DDBCCA$, we find that $P$, $T$, and $S$ are collinear. The polars of $R$ and $S$ are $AB$ and $CD$. Since $Q$ lies on the polars of $R$ and $S$, $R$ and $S$ lie on the polar of $Q$. Hence, $P$, $R$, $T$, and $S$ all lie on the polar of $Q$. $\blacksquare$ Let $X,Y$ be on the semicircle so that $B,X,Y,A$ are collinear, in that order. Let $CD$ hit $AB$ at $Z$, and let $AC$ and $BD$ intersect the semicircle at $U$ and $V$, respectively. By Pascals' theorem on $CUVDDC$, we find that $U$, $V$, and $Z$ are collinear. By the lemma, the polar of $Z$ passes through $E$. Let $XC$ hit $YD$ at $Q$, and let $XD$ hit $YC$ at $R$. $YC \perp XQ$ and $XD \perp QY$, so $R$ is the orthocenter of $\triangle QXY$, whence $QR \perp BY$. Furthermore, the polar of $Z$ also passes through $P$, $Q$ and $R$, so by the lemma, $P$, $Q$, $R$, and $E$ are collinear. It follows that $PF$ is an altitude of $\triangle PBA$, so by Blanchet's theorem $\angle CFE = \angle DFE$.

Let $AB\cap{CD}=X$,$AD\cap{BC}=Y$,$YE\cap{CD}=G$. Then $(XG;DC)=-1$,so $G$ is on the polar of $X$.As $X$ is on the $Y$'s polar $CD$,we have $YG$ is the polar of X,so $YE\perp{AB}$,we find $Y,E,F$ are collinear. Since $F(XG;DC)=-1$ and $EF\perp{AB}$,so $EF$ bisects $\angle{CFD}$.

Zhero wrote: Lemma: In cyclic quadrilateral $ABCD$, let $AD \cap BC = P$, $AB \cap CD = Q$, $AA \cap BB = R$, $CC \cap DD = S$, and $AC \cap BD = T$. Then $P$, $R$, and $S$ each lie on the polar of $Q$. Proof: By Pascal's theorem on $AACBBD$, we find that $P$, $R$, and $S$ are collinear. By Pascal's theorem on $DDBCCA$, we find that $P$, $T$, and $S$ are collinear. The polars of $R$ and $S$ are $AB$ and $CD$. Since $Q$ lies on the polars of $R$ and $S$, $R$ and $S$ lie on the polar of $Q$. Hence, $P$, $R$, $T$, and $S$ all lie on the polar of $Q$. $\blacksquare$ Typo: Pascal on AACBBD gives P, T, R collinear, instead of P, R, S. So a faster way to finish using this lemma and the fact that Zhero derived about the polar of Z passing through E is that since AB passes through the center of the circle and Z is on AB, then PE is perpendicular to AB, so line PE = line PF = line EF, so P, E, F are collinear, and we can apply Blanchet's Theorem to finish, or cyclic quadrilaterals. I saw on blackbelt14253's blog that he had used Brianchon's theorem on hexagon PCBP'AD where P' is the reflection of P across AB, which is probably a more convenient way of deducing that P, E, F collinear.

LEMMA: Suppose we have the triangle ABC and we have h_a Let h_a intersect BC at H Let P be a point on h_a Let BP intersect AC at E Let CP intersect AB at F Then:h_a bisects the angle EHF Solution: Let BC intersect AD at X From X draw a perpendicular line to AB Let's say this line intersects AB at F It's enough to prove that XF, BD, AC meet at E We do this by using Seva Now it's enough to prove that CB/FB=AD/AF Draw the line OC XFOC is cyclic --> these angles are equal COB=BXF Since OCB=XFB=90 and by using the equality above we have CB/BF=OC/XF Similarly we get AD/AF=OD/XF Since OC=OD=R WE ARE DONE

Could you please describe how you showed that $XOFD$ is cyclic? Thanks. =) EDIT: Thanks.

Dear Mathlinkers, this problem was proposed at the 35th OIM (Hongkong) but no selectde by the jury. Sincerely Jean-Louis

Let $O$ be the center of the circle. Let the tangents meet at $P$. Let H be the foot of the perpendicular from $P$ to $AB$. The key is that $AC$ and $BD$ intersect on $AH$. $PAH$ and $OAD$ are similar, so $AH/AD = HP/DO$. $PBH$ and $OBC$ are similar, so $BH/BC = HP/CO$. But $BO = CO$, so $BH/BC = AH/AD$. Also $PC = PD$ (equal tangents), so $(AH/HB)(BC/CP)(PD/DA) = 1$. Hence by Ceva's theorem applied to the triangle $ABC$, the lines $AC$, $BD$ and $PH$ are concurrent. So $H = F$. $\angle PDO = \angle PFO = \angle PCO$, so $D$, $F$ and $O$ lie on the circle diameter $PO$. So $\angle DFP = \angle DOP = \angle COP = \angle CFP$. $\Box$

let $ BC $ and $ DA $ intersect at $ G $, $ O $ is the center of the circle. connect $ GO $, $ CO $, $ DA $. try to prove $ G $, $ E $, $ F $ are conliear. in $ \Delta BGA $, use ceva theorem, try to prove: $ \frac {GC} {BC} \cdot \frac {BF} {FA} \cdot \frac {DA} {GD}=1 $. because: $ BF = BG \cdot Sin \angle GBA $, $ FA = GA \cdot Sin \angle GAB $, suppose the radium of circle $ O $ is $ R $, $ BC = \frac {R} {\tan \angle GBA} $, $DA = \frac {R}{\tan \angle GAB} $, $ GB \cdot Sin \angle GBA = GF = GA \cdot Sin \angle GAB $. so we can find $ G $, $ E $, $ F $ are conliear. $ \angle GCO = \angle GFO = 90 $, so $ G C F O $ are concylic, so $ \angle CFG = \angle COG $. $ \angle GFO = \angle GDO = 90 $, so $ G F O D $ are concylic, so $ \angle GFD = \angle GOD $. it is easy to konw $ \angle COG = \angle GOD $. prove is done.

Dear Mathlinkers, a symmetrization of the figure wrt AB, leads to another proof... without calculation... Sincerely Jean-Louis

omar31415 wrote:

omar31415,your solution is very beautiful!!!

A diagram for posterity: [asy][asy] size(10cm); pointpen = black; pair C = Drawing("C", dir(130), dir(130)); pair D = Drawing("D", dir(60), dir(60)); pair B = Drawing("B", -1.55572, dir(180)); pair A = Drawing("A", 2.0000000, dir(0)); draw(unitcircle); pair Q = Drawing(" ", extension(B,C,A,D), dir(0)); draw(B--Q); draw(A--Q); draw(A--B); pair O = Drawing(" ", 0.0, dir(0)); pair E = Drawing(" ", extension(A,C,B,D), dir(0)); pair F = Drawing(" ", extension(Q,E,A,B), dir(0)); label("$O$", O, SE); label("$F$", F, SW); label("$E$", E, SE); label("$Q$", Q, N); draw(Q--F); draw(C--A); draw(B--D); draw(circumcircle(C,Q,D), green+dashed); draw(C--F, orange+dashed); draw(C--O, red+dashed); draw(D--F, orange+dashed); draw(D--O, red+dashed); draw(D--C, red+dashed); [/asy][/asy] Let $Q=CB\cap AD$, the circle be $\Omega$, and $O$ the center of $\Omega$. Lemma: The points $Q,E,F$ are collinear. Proof: Put down Cartesian coordianates. Let $O=(0,0)$, let $\Omega$ be the unit circle, and let $AB$ be the $x$ axis. Let $C=(c_1,c_2),D=(d_1,d_2),E=(e_1,e_2),Q=(q_1,q_2)$ where $c_1^2+c_2^2=1,d_1^2+d_2^2=1$. It suffices to show that $e_1=q_1$. By similar triangles, $B=(\tfrac{1}{c_1},0)$ and $A=(\tfrac{1}{d_1},0)$. Then line $CA$ has equation \[y=\tfrac{c_2}{c_1-\tfrac{1}{d_1}}(x-\tfrac{1}{d_1}).\]Similarly, $DB$ has equation \[y=\tfrac{d_2}{1-\tfrac{1}{c_1}}(x-\tfrac{1}{c_1}).\]Since $AC\cap BD=E$, we have \begin{align*} \tfrac{c_2}{c_1-\tfrac{1}{d_1}}(e_1-\tfrac{1}{d_1})&=\tfrac{d_2}{1-\tfrac{1}{c_1}}(e_1-\tfrac{1}{c_1}) \\ \left(\tfrac{c_2}{c_1-\tfrac{1}{d_1}}-\tfrac{d_2}{d_1-\tfrac{1}{c_1}}\right)e_1&=(\tfrac{c_2}{c_1d_1-1}-\tfrac{d_2}{c_1d_1-1}) \\ e_1&=\tfrac{\tfrac{c_2-d_2}{c_1d_1-1}}{\tfrac{c_2d_1-d_2c_1}{c_1d_1-1}} \\ &=\tfrac{c_2-d_2}{d_1c_2-c_1d_2}.\end{align*}Now, line $CB$ has equation \[y=\tfrac{c_2}{c_1-\tfrac{1}{c_1}}(x-\tfrac{1}{c_1})\]and similarly line $AD$ has equation \[y=\tfrac{d_2}{d_1-\tfrac{1}{d_1}}(x-\tfrac{1}{d_1}).\]Since $AD\cap CB=Q$, we have \begin{align*} \left(\tfrac{c_1}{c_1-\tfrac{1}{c_1}}-\tfrac{d_2}{d_1-\tfrac{1}{d_1}}\right)q_1&=(\tfrac{c_2}{c_1^2-1}+\tfrac{d_2}{1-d_1^2}) \\ (\tfrac{c_1c_2}{c_1^2-1}-\tfrac{d_1d_2}{d_1^2-1})q_1&=(\tfrac{c_2}{-c_2^2}+\tfrac{d_2}{d_2^2}) \\ (\tfrac{d_1}{d_2}-\tfrac{c_1}{c_2})q_1&=(\tfrac{1}{d_2}-\tfrac{1}{c_2}) \\ q_1&=\tfrac{\tfrac{c_2-d_2}{c_2d_2}}{\tfrac{d_1c_2-d_2c_1}{c_2d_2}} \\ &=\tfrac{c_2-d_2}{d_1c_2-d_2c_1} \\ &=e_1,\end{align*}as desired. $\blacksquare$ Now we claim $FODQC$ is cyclic. Obviously $ODQC$ is cyclic since $\angle OCQ+\angle ODQ =\tfrac{\pi}{2}+\tfrac{\pi}{2}=\pi.$ Then since \[\angle OFQ=\angle OFE =\tfrac{\pi}{2}=\angle OCQ=\angle ODQ,\]$F$ lies on the same circle. Finally, we want \[\angle CFE=\angle EFD\iff\angle CFQ=\angle QFD\iff\angle QDC=\angle QCD,\]which is obvious by equal tangents. $\square$

By JBMO 2000 we have that $X,E,F$ are collinear where ${X}=AD \cap BC$. It suffices to prove that $XF$ bisects $\angle DFC$. Let $Y=AB \cap CD$ and let $Z=XF \cap CD$. Then $(Y,Z ; D,C)=-1$ and $\angle YFZ=90^{o}$ from this two facts we have that $ZF$ bisects $\angle DFC$ thus $XF$ bisects $\angle DFC$ q.e.d

Since $\angle EFO=\angle EDO=90$ and $\angle EFO=\angle ECF=90$ so we obtain E,F,O,D,c cyclic.$\angle COD=2\angle CDE$ and we have $\angle COE=\angle CDE$ so $EO$ bisects $\angle COD$ and we can obtain by cyclicity that $\angle EFD=\angle EOD=\angle COE=\angle CFE$ as desired.

The Problem can be restated as: Equivalent Problem wrote: In $\Delta ABC$, Let $\Delta DEF$ be the orthic triangle WRT $\Delta ABC$. Let tangents at $F,E$ to $\odot (BFEC)$ intersect $BC$ at $X,Y$. Let $XE,FY$ meet at $R$. Let $V$ be the foot from $R$ to $BC$.Prove, $RV$ bisects $\angle FVE$ Solution: Let $H$ be the orthocenter of $\Delta ABC$ and $M$ be the midpoint of $AH$. Then, $M$ lies on $XF, YE$. Let the perpendiculars from $E, F$ to $BC$ intersect $XM,YM$ at $Q,P$. Let $PF,QE$ intersect $\odot (BFEC)$ again at $F', E'$ $\implies$ $FE,BC,E'F'$ concur, suppose say at $T$, then, Apply Pascal on $FFEEE'F'$ $\implies$ $PQ$ passes through $T$. By La Hire's Theorem $\implies$ $XE,FY$ are polars of $P,Q$ WRT $\odot (BFEC)$. Now since, $T,P,Q$ are collinear, their polars are concurrent $\implies$ $XE \cap FY$ $=$ $R$ lies on $AH$ $\implies$ $V \equiv D$ and since $H$ is the incenter of $\Delta DEF$ $\implies$ $RD$ ($RV$) bisects $\angle FDE$ ($\angle FVE$)

So let's define point $P = AD \cap BC$ Let point $Q \in AB$ such that $PQ \perp AB$ Then the pentagon $CQODP$ is cyclic. Then we have 2 similarity relations $\triangle PAQ \sim \triangle OAD$ and $\triangle PBQ \sim \triangle OBC $ Thus we have the following relation: $$ \frac{AD}{AQ} = \frac{OD}{PQ} = \frac{OC}{PQ} = \frac{BC}{BQ} $$Thus we have the following: $$ \frac{AQ}{QB}\frac{BC}{CP}\frac{PD}{DA} = 1 $$Because $PD = PC$ and because of the ratios above. Thus by the reverse Ceva theorem we have that $AC,BD$ and $PQ$ are concurrent cevians,thus $Q \equiv F$ Thus we have that $CFODP$ is cyclic This angle-chase finishes the problem off: $$\angle DFP = \angle DOP = \angle POC = \angle PFC$$Thus we have shown,because $\angle DFP = \angle PFC$, that PF is an angle-bisector of the angle $\angle CFD$

Let $W$ be the intersection of $BC$ and $DA$. Let $W’$ be the foot of the perpendicular from $W$ onto $AB$. Let $O$ be the center of the given semicircle. Then, since $\triangle WBW’ \sim \triangle BCO$, $$\frac{OC}{BC} = \frac{WW’}{BW'}$$Since $\triangle WW’A \sim \triangle ODA$, $$\frac{DA}{OD} = \frac{W'A}{WW'}$$Multiplying, we see that $\frac{BC}{DA} = \frac{W'A}{BW’}$. We also know that $WC = WD$ by power of a point. Therefore, by Ceva’s theorem, we see that $CA,BD$, and $WW’$ are concurrent. Furthermore, we know that $CWDOW’$ is cyclic because $\angle WCO = \angle WDO = \angle WW’O = 90$. Now, we proceed with angle-chasing. Let $\angle WW’D = a$. Then, $\angle WOD = \angle WCD = a$. Then, $\angle DW’O = \angle DCO = \angle OWD = 90-a$. Since $O$ is the center of the semicircle, $\angle CDO = 90-a$ as well, so $\angle WOC = a$.Thus, $WO$ bisects $\angle COD$, which as we showed beforehand is the same as saying that $EF$ bisects $\angle CFD$.

@above I think you misinterpret the problem. Read carefully definition of point $E$. My solution: Let $X = AD \cap BC$ and $Y,P$ - reflections of $X,C$ about $AB$. By Brianchon on $ADXCBY$ we see that $E \in XY$ and so $F \in XY$ By Brianchon on $ADXBPY$ we see that $F \in DP$ which implies the result.

Really similar to post #7. Oh well. The key observation is that $AD,BC,EF$ concur. Let $AD$ and $BC$ intersect at $P$ and let $Q$ be the foot of the altitude from $P$ to $AB.$ Also let the semicircle have center $O.$ Now note \[\triangle PAQ\sim \triangle OAD\]\[\triangle PBQ\sim \triangle OBC\]so $\frac{AQ}{QB}\cdot\frac{BC}{CP}\cdot\frac{PD}{DA}=1.$ Since $AC,BD,PQ$ concur, $Q$ is actually $F,$ and $AC,BD,PF$ concur. Now note \[\angle OCP=\angle ODP=\angle OFP=90^{\circ},\]so $OFCPD$ is cyclic. Thus \[\angle COP=\angle DOP\]\[\angle CFP=\angle DFP.\]

Let A' be the point on the diameter for semicircle closer to B. It is not hard to see, with a simple diagram (drawing perp. angles are easy), that we can suspect concurrency. If the intersection of AD and BC is G, by similar triangles ADO with AFG and BCO with BFG, we have by Ceva's that they (AD, BC, and EF) concur. <ODG=<OCG=<OFG=90 degrees, so OFCPD is cyclic. We then have <CFG=<CDG=<CA'D, and <DFG=<DCG=<DA'C, where the last two steps follow from the well known tangent angle rule (or the fact that it is an "instantaneous same point" that are angle <DCG=<DCC=DA'C). Edit: Will attach ggb diagram later