mr.danh wrote: The incircle of triangle $\triangle ABC$ touches the sides $BC$, $CA$, $AB$ at $D, E, F$ respectively. $X$ is a point inside triangle of $\triangle ABC$ such that the incircle of triangle $\triangle XBC$ touches $BC$ at $D$, and touches $CX$ and $XB$ at $Y$ and $Z$ respectively. Show that $E, F, Z, Y$ are concyclic. Proof: Since $BZ = BD = BF$ and $CY = CD = CE$,we conclude that $\frac {BF}{CE} = \frac {BZ}{CY}$,hence $BC,EF,ZY$ are concurrent,and let $T$,be their intersection point.As we know $TD^2 = TE\cdot TF$,and $TD^2 = TZ\cdot TY$,hence $TE\cdot TF = TZ\cdot TY$,so $E,F,Z,Y$ are concyclic.

This comes from the IMO Shortlist from 1995. See also here: http://www.mathlinks.ro/viewtopic.php?t=37621.

Try inverting in D and the problem gets trivial

Proof taken from grobber: Let's prove the following lemma: With the exact same notations, if $T = EF\ap BC$, then $T$ is the harmonic conjugate of $D$ wrt $B,C$. Let $P = EF\cap AD$. $T$ lies on the polar of $A$, so $A$ lies on the polar of $T$. At the same time, $D$ lies on the polar of $T$, so the polar of $T$ is $AD$ (I mean polar wrt the incircle). Since $P\in TEF$ and $P$ belongs to the polar of $T$, it means that $(T,P;F,E) = - 1$. However, $(T,P;F,E) = (T,D;B,C)$, so we get the conclusion: $(T,D;B,C) = - 1$. The lemma is proven. We apply the lemma to $ABC$ and $XBC$ to find that $T = EF\cap BC$ and $T' = YZ\cap BC$ coincide (they must both be equal to the harmonic conjugate of $D$ wrt $BC$). Now $TE\cdot TF = TD^{2} = TY\cdot TZ$, and we are done.

Let $\angle ABC = 2\beta, \angle ZBD = 2x, \angle BCA = 2\theta, \angle YCD = 2y$. So $\angle YZD = 90^{\circ}-y$ and $\angle FED = 90^{\circ}-\beta$. Since $BF=BZ=BD$, $Z$ is on the circle with center $B$ and radius $[BD]$. Thus $\angle FZD = 180^{\circ} + \beta \Rightarrow \angle FZY = 90^{\circ}+\beta + y$. Similarly $\angle YED = y$ and $\angle FEY = \angle 90^{\circ} - \beta - y$. So $\angle FZY + \angle FEY = 180^{\circ}$. This implies $F,E,Y,Z$ are concyclic.

Dear Mathlinkers, 1. EF, YZ and BC are concurrent (prove before) 2. According to Monge's theorem we are done. Sincerely Jean-Louis

excuse me I dont understand why $EF$ ; $ZY$ ; $BC$ are concurrent . Please explain more .

siavosh wrote: excuse me I dont understand why $EF$ ; $ZY$ ; $BC$ are concurrent . Please explain more . siavosh, $EF$ and $BC$ concur at a point $T$. Now, I don't know if you are familiar with harmonic division, but if you are, then it won't be hard to understand that as $(TBDC)$ (or $(TCDB)$, on the relative positions of $B$ and $C$) is harmonic (because $AD$, $BE$ and $FC$ are concurrent), and $P=BC\cap ZY$ is such that $(PBDC)$ is harmonic (because $BY$, $CZ$ and $DX$ concur), than $P\equiv T$. I hope you'll understand it now. Regards

In order to prove that $EF,YZ,BC$ are concurrent, you can use Menelaus' Theorem on $\triangle ABC,\triangle XBC.$, as we can easily obtain $BF=BY,CE=CZ$. The rest is easy.

Let $P=FE \cap BC$ Sine rule in $\triangle{BFB}$ gives $\frac{PB}{cos\frac{A}{2}}=\frac{s-b}{sin{\angle{BPF}}}$ Similarly sine rule in $\triangle{PEC}$ gives $\frac{PC}{cos\frac{A}{2}}=\frac{s-c}{sin{\angle{CPE}}}$ Dividing these two relations we get $\frac{PB}{PC}=\frac{s-b}{s-c}$ Thus in $\triangle{BXC}$ we have $\frac{PB}{PC} \frac{CY}{XY} \frac{XZ}{YZ}=\frac{s-b}{s-c} \frac{s-c}{XY} \frac{XZ}{s-b} =1$ so by converse of menelaus theorem,$P,Z.Y$ are collinear. Thus $PF*PE=PZ*PY=PD^2$ and we are done!! Also note that $P$ is the radical center of $\odot{DEF},\odot{DYZ}$ and $\odot{EFZY}$.

Obviously $AB,BC,CA$ are tangent to their incircle and $BX,CX,CB$ tangent to theirs. Then $BF=BD=BZ$ and $CE=CD=CY$. We angle chase: $$\begin{align*} \angle EFD+\angle EYZ &= \angle EFD+\angle DFZ+\angle DYZ+\angle EYD \\ &=\angle EFD+\angle DFB-\angle BFZ+\angle DYZ+\angle DYC-\angle EYC \\ &=\angle EFD +\angle DEF -\tfrac{\pi -\angle FBZ}{2}+\angle DYZ+\angle DZY-\tfrac{\pi -\angle ECY}{2} \\ &=\pi -\angle FDE +\pi -\angle ZDY-\pi +\tfrac{\angle FBD+\angle DBZ+\angle ECD+\angle DCY}{2} \\ &=\pi -\angle FDE -\angle ZDY +\tfrac{\pi -2\angle BFD +\pi -2\angle BDZ +\pi -2\angle DEC +\pi-2\angle CDY}{2} \\ &=3\pi -\angle FDE -\angle ZDY -(\angle FED +\angle DYZ+\angle DFE+\angle DZY) \\ &= 3\pi -\angle FDE-\angle ZDY -(\pi -\angle FDE +\pi -\angle ZDY) \\ &= \pi , \end{align*}$$ so $EFZY$ is cyclic as desired. $\blacksquare$ EDIT: An additional implied result is that $\triangle DZY$ is isosceles with $DZ=DY$!

Let $BC$ intersect $EF$ at $P$, and let $ZY$ intersect $BC$ at $P'$. Now $AD,BE,CF$ are concurrent, and so $(B,C;D,P)=-1$. Again, $XD,BY,CZ$ are concurrent, and thus $(B,C;D,P)=-1$. So we must have $P\equiv P'$. Thus $BC,ZY,FE$ are concurrent. Now $PE\cdot PF=PD^{2}=PC\cdot PB$ and we are done.

There is also an interesting result, which is implicated by the thesis of this problem. Namely, notice that if $P=EY \cap FZ$ and $S= DP \cap EF$, then $AS$ is a bisector of $BC$.

mr.danh wrote: The incircle of triangle $\triangle ABC$ touches the sides $BC$, $CA$, $AB$ at $D, E, F$ respectively. $X$ is a point inside triangle of $\triangle ABC$ such that the incircle of triangle $\triangle XBC$ touches $BC$ at $D$, and touches $CX$ and $XB$ at $Y$ and $Z$ respectively. Show that $E, F, Z, Y$ are concyclic. My solution. Let $P$ be the intersection of $BC$ and $EF.$ Because of the concurency of the lines $AD$,$BE$,$CF$ in the Gergonne' point of triangle $ABC$, we deduce by using Ceva' and Menelaus' theorems in the triangle $ABC$ that, $$\frac{\overline{BP}}{\overline{PC}}\cdot\frac{\overline{CD}}{\overline{DB}}=-1.$$ which means that the cross ratio $\left(B,C,P,D \right)=-1$, that is $\left(B,C,P,D \right)$ is harmonic bundle. A similar argument shows that, $\left(B,C,Q,D \right)$ is harmonic bundle, where $Q$ is the intersection of $YZ$ and $BC.$ Now, it is easy to see that, $Q$ and $P$ are the same!, which means that the lines, $EF$,$YZ$, and $BC$ are concurrent at $P.$ Therefore, we have $$\overline{PY} \cdot \overline{PZ}=\mathcal{P}_{\odot YZD}(P)=PD^2=\mathcal{P}_{\odot EFD}(P)=\overline{PF} \cdot \overline{PE}$$ where $\mathcal{P}_{\omega}(A)$ denote the power of point $A$ with respect to the circle $\omega.$ This prove that the points $E, F, Z, Y$ are concyclic.

Other solution: Perpendicular bisectors of $EF$, $FZ$, $ZY$, $YF$ are the angle bisectors of $\angle BAX$, $\angle ACB$, $\angle CXB$, $\angle XBA$. $ZYFE$ is cyclic iff bisectors of $EF$, $FZ$, $ZY$, $YF$ are concurrent iff bisectors of $\angle BAX$, $\angle ACB$, $\angle CXB$, $\angle XBA$ are concyclic iff $ABXC$ is circumscribed, what can be easily proven by calculating it's side lengths.

Isn't this trivial by this following (well known ?) lemma ? (Denoting $\omega_X^{Y}$ by the circle centered at $X$ passing through $Y$) Lemma: If four circles $\omega_{A_1}^{O}, \omega_{A_2}^{O}, \omega_{A_3}^{O}, \omega_{A_4}^{O}$ satisfy the relation that $\omega_{A_i}^{O}$ is tangent to $\omega_{A_i+2}^{O}$ at $O$, and if $A_1A_3 \perp_O A_2A_4$, then the points $\omega_{A_i}^{O} \cap \omega_{A_i+1}^{O}$ are concylic.

We can draw a circle $\omega_1$ centered at $B$ passing through $F,Y,D$ and a circle $\omega_2$ centered at $C$ passing through $E,Z,D$. Upon inverting through $D$, $w_1$ and $w_2$ both become lines perpendicular to $\overleftrightarrow{BC}$, and $E'F'Y'Z'$ becomes a rectangle. In particular, it is cyclic, so $EFYZ$ is cyclic.

IMO Shortlist 1995, G3 wrote: The incircle of triangle $\triangle ABC$ touches the sides $BC$, $CA$, $AB$ at $D, E, F$ respectively. $X$ is a point inside triangle of $\triangle ABC$ such that the incircle of triangle $\triangle XBC$ touches $BC$ at $D$, and touches $CX$ and $XB$ at $Y$ and $Z$ respectively. Show that $E, F, Z, Y$ are concyclic. Solution:- Let $YZ\cap BC=K$ and $EF\cap BC=K'$. We will try to show that $K'\equiv K$. Claim 1:- $(K,D;B,C)=-1$. Note that $XD,YB,CZ$ concur at the Gregonne Point of $\triangle BXC$. Hence, $(K,D;B,C)=-1$. Claim 2:- $(K',D;B,C)=-1$. Again notice that $AD,BE,CF$ concur at the Gregonne Point of $\triangle ABC$. Hence, $(K',D;B,C)=-1$. Now as $-1=(K,D;B,C)=(K',D;B,C)\implies K'\equiv K$. Hence, $EF,ZY,BC$ concurs at a point $K$. Hence, $$KF.KE=KD^2=KZ.KY\implies E,F,Z,Y\text{ are concyclic}.\blacksquare$$ Another Solution Draw a circle centered at $B$ passing through $F,Z,D$ and another circle centered at $C$ passing through $E,Y,D$. Now the problem is same as this one. EGMO 8.23 . $\blacksquare$.

[asy][asy] size(11cm); defaultpen(fontsize(10pt)); pair A = dir(140), B = dir(210), C = dir(330), I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,C,A), F = foot(I,A,B), P = extension(B,C,E,F), I1 = I+dir(I--D)*abs(I-D)*0.35; path p = circle(I1,abs(D-I1)); pair Z = intersectionpoints(circle(B,abs(B-D)),p)[1], Y = intersectionpoints(circle(C,abs(C-D)),p)[0], X = extension(B,Z,C,Y); dot("$A$", A, dir(120)); dot("$B$", B, dir(270)); dot("$C$", C, dir(330)); dot("$D$", D, dir(270)); dot("$E$", E, dir(15)); dot("$F$", F, dir(135)); dot("$P (Q)$", P, dir(210)); dot("$X$", X, dir(100)); dot("$Y$", Y, dir(260)); dot("$Z$", Z, dir(315)); draw(C--B--A--C, linewidth(1.2)); draw(B--X--C); draw(Y--P--E^^P--B, dashed); draw(incircle(A,B,C)^^p); draw(circumcircle(E,F,Y), dotted); [/asy][/asy] If $AB = AC$ then $EFZY$ is just an isosceles trapezoid. Othersise, let $P = \overline{EF} \cap \overline{CB}$ and $Q = \overline{YZ} \cap \overline{BC}$. Note that $P$ and $Q$ must lie on the same side of segment $\overline{BC}$. We claim that $P = Q$. Indeed, by Menelaus's Theorem on triangles $ABC$ and $XBC$, we have $$\dfrac{AF}{FB} \cdot \dfrac{BP}{PC} \cdot \dfrac{CE}{EA} = \dfrac{XZ}{ZB} \cdot \dfrac{BQ}{QC} \cdot \dfrac{CY}{YX} = 1.$$ We know that $AF = AE$, $XZ = XY$, $BF = BZ = BD$, and $CD = CY = CE$, so the above equation simplifies to $$\dfrac{BP}{PC} = \dfrac{BQ}{QC},$$ implying $P = Q$. Finally, we have $$PZ \cdot PY = PD^2 = PF \cdot PE$$ by Power of a Point, and the result follows. $\blacksquare$

IMO Shortlist 1995 G3 wrote: The incircle of triangle $\triangle ABC$ touches the sides $BC$, $CA$, $AB$ at $D, E, F$ respectively. $X$ is a point inside triangle of $\triangle ABC$ such that the incircle of triangle $\triangle XBC$ touches $BC$ at $D$, and touches $CX$ and $XB$ at $Y$ and $Z$ respectively. Show that $E, F, Z, Y$ are concyclic. $AD, BE, CF$ concur at the Gergonne Point of $\triangle ABC \Longrightarrow (EF \cap BC, D;B,C) = -1$ $XD, BY, CZ$ concur at the Gergonne Point of $\triangle XBC \Longrightarrow (YZ \cap BC, D; B,C) = -1$ This forces $EF \cap BC \equiv YZ \cap BC$ or $EF, YZ,BC$ concur(say at a point $P$). Thus, $PD^2 = PZ \cdot PY = PF \cdot PE$, and we are done by Converse of Power of a Point Theorem.

Storage. Note that $AD,BE,CF$ and $XD,BY,CZ$ are concurrent at Gergonne point of $\triangle ABC,\triangle XBC$, respectively. Thus, if $P_1=EF\cap BC$ and $P_2=ZY\cap BC$, we have $$(P_1,D,B,C)=-1=(P_2,D,B,C)\implies P_1\equiv P_2(\equiv P),$$ hence $EF,YZ,BC$ are concurrent. Now, we conclude by PoP, $PE\cdot PF=PD^2=PZ\cdot PY$, we are done.

$- AD,BE,CF$ are concurrent $\implies (C,B;D,T)=-1$ $-XD,CY,BZ$ are concurrent $\implies (C,B;D,YZ \cap BC)=-1$ Then: $T= YZ\cap BC$ $\implies PZ.PY = PD^2 = PE.PF \implies E, F, Z, Y$ are concyclic.

Let $\overline{BC}$ intersect $\overline{EF}$ and $\overline{YZ}$ at $W$ and $W',$ respectively. Since $$(B,C;D,W)=(B,C;D,W')=-1,$$ $W=W'.$ Hence, $$WE\cdot WF=WD^2=WY\cdot WZ.$$ $\square$

Let $YZ\cap BC=S$. Since $XD,BY,CZ$ are concurrent (at the Gergonne Point), we have $(S,D;B,C)=-1$. Similarly, let $EF\cup BC=S'$, we have $(S',D;B,C)=-1\implies S\equiv S'$. Note that $Pow(S,\text{incircle}(BXC))=SZ\cdot SY = SD^2$. Also, $Pow(S,\text{incircle}(ABC))=SF\cdot SE=SD^2$. Thus, $SZ\cdot SY=SF\cdot SE$ so $E,F,Z,Y$ are concyclic, as desired.

Applying Ceva-Menelaus on the Gergonne points of $ABC$ and $XBC$ implies $BC, EF, YZ$ are concurrent at some point $T$. Thus, we have $$TE \cdot TF = | Pow_{(DEF)}(T) | = TD^2 = | Pow_{(DYZ)}(T) | = TY \cdot TZ$$ which clearly finishes. $\blacksquare$ Remark: This problem is extremely headsolveable.

Let $T_1= \overline{EF} \cap \overline{BC}$ and $T_2 = \overline{ZY} \cap \overline{BC}$. As $$(T_1D;BC) = (T_2D;BC) = -1$$ the lines $\overline{EF}, \overline{ZY}, \overline{BC}$ are concurrent at $T_1 = T_2$. Thus $EFZY$ is cyclic by radical axis.

Erken wrote: mr.danh wrote: The incircle of triangle $\triangle ABC$ touches the sides $BC$, $CA$, $AB$ at $D, E, F$ respectively. $X$ is a point inside triangle of $\triangle ABC$ such that the incircle of triangle $\triangle XBC$ touches $BC$ at $D$, and touches $CX$ and $XB$ at $Y$ and $Z$ respectively. Show that $E, F, Z, Y$ are concyclic. Proof: Since $BZ = BD = BF$ and $CY = CD = CE$,we conclude that $\frac {BF}{CE} = \frac {BZ}{CY}$,hence $BC,EF,ZY$ are concurrent,and let $T$,be their intersection point.As we know $TD^2 = TE\cdot TF$,and $TD^2 = TZ\cdot TY$,hence $TE\cdot TF = TZ\cdot TY$,so $E,F,Z,Y$ are concyclic. $BZ$ And $BD$ are different

Really easy for a G3

Since we know that $AD, BE$ and $CF$ concur at a point.(Gergonne point) Let $EF$ meet $BC$ at $K.$ We know that $(K,D;B,C)=-1$ . (This directly comes from applying menelaus and ceva's) Similarly let $YZ$intersect $BC$ at $K',$ but since $(K',D;B,C)=-1$ $\implies K=K' \implies EF$ and $YZ$ intersect on $BC$ at $K.$ since $KY*KZ=KD^2=KF*KE$ $\implies$ $E, F, Z, Y$ are concyclic.

Note that because of the Ceva-Menelaus configuration and the fact that the intouch cevians concur, we have that $EF$ and $YZ$ both meet $BC$ at the harmonic conjugate $T$ of $D$ with respect to $B, C$, i. e. the point $T$ such that $(B, C; D, T) = -1$. In particular, $EF, YZ, BC$ concur. Now we have $TE \cdot TF = TD^2 = TY \cdot TZ$ by Power of a Point, which implies the result. $\square$