We know that the orthocenter reflects over the sides of the triangle on the circumcircle. Therefore the minimal distance $ OD+HD$ equals $ R$. Obviously we can achieve this on all sides, so we assume that $ D,E,F$ are the intersection points between $ A',B',C'$ the reflections of $ H$ across $ BC,CA,AB$ respectively. All we have to prove is that $ AD$, $ BE$ and $ CF$ are concurrent. In order to do that we need the ratios $ \dfrac {BD}{DC}$, $ \dfrac {CE}{EA}$ and $ \dfrac {AF}{FB}$, and then we can apply Ceva's theorem. We know that the triangle $ ABC$ is acute, so $ \angle BAH = 90^\circ- \angle B = \angle OAC$, therefore $ \angle HAO = |\angle A - 2(90^\circ -\angle B)| = |\angle B- \angle C|$. In particular this means that $ \angle OA'H = |\angle B-\angle C|$. Since $ \angle BA'A = \angle C$ and $ \angle AA'C = \angle B$, we have that $ \angle BA'D = \angle B$ and $ \angle DA'C = \angle C$. By the Sine theorem in the triangles $ BA'D$ and $ DA'C$, we get \[ \dfrac {BD}{DC} = \dfrac { \sin B }{\sin C }.\] Using the similar relationships for $ \dfrac {CE}{EA}$ and $ \dfrac {AF}{FB}$ we get that those three fractions multiply up to 1, and thus by Ceva's, the lines $ AD, BE$ and $ CF$ are concurrent.

Lemma 1: If an ellipse is inscribed triangle $ABC$, tangent to $BC$, $AC$, and $AB$ at $D$, $E$, and $F$, respectively, then $AD$, $BE$, and $CF$ are concurrent. Proof: Scale the ellipse about its major axis so that it becomes a circle; here, $A'D'$, $B'E'$, and $C'F'$ concur at the Gergonne point of $A' B' C'$. Scaling preserves incidence, so $AD$, $BE$, and $CF$ concur. Lemma 2: Let $\ell$ be any line, and let $X$ and $Y$ be any points on the same side of $\ell$. Let $Z$ be the reflection of $X$ across $\ell$, and let $W$ be the intersection of $YZ$ and $\ell$. Then the ellipse with foci $X$ and $Y$ that passes through $W$ is tangent to $\ell$. Proof: Suppose that the ellipse intersected line $\ell$ at another point, $W'$. Then $XW' + YW' = YW' + ZW' > YZ$ by the triangle inequality, since $W'$, $Y$, and $Z$ are not collinear. On the other hand, by definition of ellipse, $XW' + YW' = XW + YW = ZW + YW = YZ$, so $YZ > YZ$, which is a contradiction. By lemma 1, it is sufficient to show that there exists an ellipse with foci $O$ and $H$ that are tangent to the sides $BC$, $AC$, and $AB$ at $D$, $E$, and $F$, respectively. Reflect $H$ across $BC$, $CA$, and $AC$ to get $A'$, $B'$, and $C'$, respectively, and let $OA'$ hit $BC$ at $D$, $OB'$ hit $CA$ at $E$, and $OC'$ hit $AB$ at $F$. . $A',B',C'$ lie on the circumcircle of $ABC$, so $R = OA' = OB' = OC' = HD + DO = HE + EO = HF + FO$, where $R$ is the circumradius of $\triangle ABC$ (since $A', B', C'$ are the reflections of $H$ across the sides of the triangle.) Consider the ellipse with foci $O$ and $H$ with a major axis of length $R$. By definition, the ellipse passes through $D$, $E$, and $F$. However, by lemma 2, it is tangent to sides $BC$, $CA$, and $AB$, so by lemma 1, we are done.

@Zhero: Instead of scaling, it is more elegant to just use Brainchon's theorem. Cheers, Rofler

Probably. I believe we could also just centrally project it to a circle as well (my first approach). I chose to scale because I felt it was the most elementary solution; anyone who knows anything about ellipses should be able to understand the solution I gave.

since $H$ and $O$ are two isogonal conjugate points, there exists an ellipse which these two points are its foci and its tangent to sides of the triangle.

Valentin Vornicu wrote: By the Sine theorem in the triangles $ BA'D$ and $ DA'C$, we get \[ \dfrac {BD}{DC} = \dfrac { \sin B }{\sin C }.\] By the law of sines for triangles $~$ $ \triangle{BA'D} $ $~$ and $~$ $ \triangle{DA'C} $ $~$ we get: \[ \frac { BD } { \sin B } = \frac { A'D } { \sin \angle{DBA'} } \quad \wedge \quad \frac { DC } { \sin C } = \frac { A'D } { \sin \angle{DCA'} } \] So for: \[ \dfrac {BD}{DC} = \dfrac { \sin B }{\sin C }.\] to be true, the following: \[ \sin \angle{DBA'} = \sin \angle{DCA'} \] should be true as well. I don't think that's correct.

You are right. See here PP8.

Virgil Nicula wrote: You are right. See here PP8. Wow you had a math blog!!! I unfortunately can't read the entire problems. Only half of the picture is displayed. Do you know why?

See click on the title of message.

orl wrote: Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\] and the lines $AD$, $BE$, and $CF$ are concurrent. This problem also appeared in one of the Indian TSTs!!!

Take $H_a, H_b, H_c$ be the mirror images of $H$ about $BC, AC, $ and $AB$ respectively. They lie on the circumcircle of $\Delta ABC$. Now, take $\{D\} = OH_a \cap BC$, $\{E\} = OH_b \cap AC$ and $\{F\} = OH_c \cap AB$. Now, using Pool's Theorem, we get that $OD + DH = OE + EH = OF + FH$. Moreover, in this case the sum $OD+DH$ is minimal, so the points $D, E, F$ are situated on an ellipse of foci $O$ and $H$, having the constant sum equal to $R$ (the ray of the circumcircle of $\Delta ABC$), and which is tangent to the edges of the triangle. Now, this is just Gergonne's theorem, under a projective transformation.

Here is a bashing proof. $A',B',C'$ are the reflection point of $H$ with respect to $BC,CA,AB$ respectively $D,E,F$ are the intersection point of $OA',OB',OC'$ with $BC,CA,AB$ respectively Let $ S_A : = (b^2 + c^2 - a^2) [ b^2 (c^2 + a^2 - b^2 )^2 + c^2 (a^2 +b^2 - c^2 )^2 ] $ $ S_B : = (c^2 + a^2 - b^2 ) [ c^2 (a^2 + b^2 - c^2 )^2 + a^2 (b^2 +c^2 - a^2 )^2 ]$ $ S_C : = (a^2 + b^2 - c^2 ) [ a^2 (b^2 +c^2 -a^2)^2 + b^2 (c^2 +a^2 -b^2 )^2 ] $ One can calculate the barycentric cooridnates of $D,E,,F$ respectively to be $ D = (0 : \frac{1}{S_B} : \frac{1}{S_C} ) $ $ E = (\frac{1}{S_A} : 0 : \frac{1}{S_C} ) $ $ F = (\frac{1}{S_A} : \frac{1}{S_B} : 0 ) $ It is straightforward to see that $AD,BE,CF$ are concurrent.

Wow this is an $\text{IMO}$ problem...

So, let's play with it...

zuss77 wrote: So, let's play with it... How did you do that?

How do you attach a gif?

orl wrote: Let $O$ be the circumcenter and $H$ the orthocenter of an acute triangle $ABC$. Show that there exist points $D$, $E$, and $F$ on sides $BC$, $CA$, and $AB$ respectively such that \[ OD + DH = OE + EH = OF + FH\]and the lines $AD$, $BE$, and $CF$ are concurrent.

we can draw an $elipse$ with focuses $O$ and $H$ that is tangent to the sides of $ABC$ in $D,E,F$ then from $brianchon's theorem AD,BE,CF$ concur. $\blacksquare$

When I solved this I thought of the following problem: Let $A,B$ be points in the plane and $l$ be a line in that plane. Find the locus of the point(s) $S$ such that $AS+BS$ is minimal. This is solved by Euclid's lemma and reflecting $B$ or $A$ over $l$.

Let $R$ be the circumradius of $ABC$, $AH, BH, CH$ meet $(ABC)$ again at $H_a, H_b, H_c$ respectively, and $D = BC \cap OH_a$, $E = CA \cap OH_b$, $F = AB \cap OH_c$. By Orthocenter Reflections, it's clear that $$OD + DH, OE + EH, OF + FH = R.$$It's well-known that $O$ and $H$ are isogonal conjugates. Hence, the Ratio Lemma and LoS implies $$\frac{BD}{DC} = \frac{H_aB}{H_aC} \cdot \frac{\sin BH_aD}{\sin CH_aD} = \frac{\sin H_aCB}{\sin H_aBC} \cdot \frac{\sin CH_aA}{\sin BH_aA} = \frac{\sin H_aAB}{\sin H_aAC} \cdot \frac{\sin CBA}{\sin BCA}$$$$= \frac{\sin (90^{\circ} - B)}{\sin (90^{\circ} - C)} \cdot \frac{\sin B}{\sin C} = \frac{\cos B \sin B}{\cos C \sin C}.$$Now, finding $\frac{CE}{EA}, \frac{AF}{FB}$ via analogous processes and applying Ceva's finishes easily. $\blacksquare$ Remark: This solution is clearly motivated by the Orthocenter Reflection Lemma.

Step1 : Determine where are $D,E,F$. Let $H1$ be reflection of $H$ across $BC$ and let $OH_1$ meet $BC$ at $D$. Use same trick to define $E$ and $F$ now note that: $DH + DO = DO + DH_1 = OH_1$ , $EH + EO = EO + EH_2 = OH_2$ , $FH + FO = FO + FH_3 = OH_3$ and $OH_1 = OH_2 = OH_3 = R$. Step2 : proving $AD,BE$ and $CF$ are concurrent. By Ceva's Theorem we only need to prove $\frac{AF}{FB} . \frac{BD}{DC} . \frac{CE}{EA} = 1$. $\frac{BD}{DC} = \frac{\sin{BCD}}{\sin{DBC}} . \frac{\sin{DH_1B}}{\sin{CH_1D}}$ and $\angle DH_1B = \angle HAO + \angle C = \angle B - \angle C + \angle C = \angle B$ and $\angle CH_1D = \angle B + \angle C - \angle B = \angle C$. now if we multiple all we will have 1 so $AD,BE$ and $CF$ are concurrent.

Let $H_A,H_B,$ and $H_C$ be the reflections of $H$ over $\overline{BC},\overline{AC},$ and $\overline{AB},$ respectively. Let $D=\overline{OH_A}\cap\overline{BC}$ and define $E$ and $F$ similarly. Notice $$DO+DH=DO+DH_A=OH_A=r$$where $r $ is the radius of $(ABC).$ Similarly, $EO+EH=FO+FH=r.$ We see $$\angle HAO=\angle A-2(90-\angle B)=\angle A+2\angle B-180.$$Hence, by the Law of Sines, $$\frac{BD}{CD}=\frac{\sin\angle BH_AD\cdot DH_A/\sin\angle H_ABC}{\sin\angle CH_AD\cdot DH_A/\sin\angle H_ACB}=\frac{\sin(\angle C+\angle HAO)\cos\angle B}{\sin(\angle B-\angle HAO)\cos\angle C}=\frac{\sin\angle B\cos\angle B}{\sin\angle C\cos\angle C}$$and Ceva finishes. $\square$

What is branchion's. Let $H_1,H_2,H_3$ be reflections of $H$ over $BC,AC,AB$ respectively. Let $OH_1,OH_2,OH_3$ intersect $BC,AC,AB$ at $E,F,D$ respectively. Let $r$ be the circumradius. Note that $(OD+DH,OE+EH,OF+FH)=(OD+DH_3,OE+EH_1,OF+FH_2)=(r,r,r).$ Finally, note that $D,F,E$ are collectively on an ellipse with foci $O,H.$ Since $D,E,F$ minimize the distance $OD+DH,OE+EH,OF+FH$ respectively, this ellipse must be tangent to $\triangle ABC'$s sides. Then we can do a scaling in the direction of $OH$ that makes this ellipse into a circle. This transformation does not disrupt intersections, so this ellipse is still tangent to $\triangle ABC'$s sides, and it preserves concurrentness. This ellipse is now an incircle so $AE,BF,CD$ are concurrent. I know I messed up point labeling so just substitute $D$ in for $E,$ $F$ in for $E$ and $D$ in for $F$ to get the result.

We claim that the conditions are satisfied when $D=OH_A \cap BC$ and define $E,F$ symmetrically, where $H_A$ is the reflection of $H$ over $BC$. By the reflecting the orthocenter lemma, have that $OD+DH=OD+DH_A=R$, where $R$ is the circumradius of $ABC$, so the first condition holds true. We now prove the concurrency using complex numbers. By the reflection formula, we get that $h_a=\frac{-bc}{a}$. Next using the intersection of line formulas we have $d=\frac{bc(b+c)}{a^2+bc}$. We have $\frac{BD}{DC} = \frac{\lambda}{1-\lambda}$ where $\lambda$ satisfies $\lambda b + (1-\lambda ) c = \frac{bc(b+c)}{a^2+bc}$. This gives $\lambda=\frac{c(b^2-a^2)}{(b-c)(a^2+bc)}$. Thus by Ceva's Theorem, the problem is equivalent to $$\prod_{\text{cyc}} \frac{\frac{c(b^2-a^2)}{(b-c)(a^2+bc)}}{1-\frac{c(b^2-a^2)}{(b-c)(a^2+bc)}} = \prod_{\text{cyc}} \frac{c}{b} \cdot \frac{b^2-a^2}{a^2-c^2} =1,$$but this is obvious so we are done.

Reflect $H$ over the sides $BC,AC,AB$ of the triangle to get the points $H_A,H_B,H_C$, which must lie on $(ABC)$. Note that if we let $D$ be the intersection of lines $OH_A$ and $BC$, and similarly for $E$ and $F$, then the $OH+DH=OE+EH=OF+FH$ condition is satisfied. We now apply law of sines on $\triangle BH_AD$ and $\triangle CH_AD$ and then angle chase to find that $$\frac{BD}{DC}=\frac{\sin{\angle BH_AD}}{\sin{\angle CH_AD}}\cdot \frac{\sin{\angle H_ABD}}{\sin{\angle H_ACD}}=\frac{\sin{\angle B}}{\sin{\angle C}}\cdot \frac{90^\circ-\sin{\angle B}}{90^\circ-\sin{\angle C}}.$$We get similar results for $\tfrac{CE}{EA}$ and $\tfrac{AF}{FB}$, so $\tfrac{BD}{DC}\cdot\tfrac{CE}{EA}\cdot \tfrac{AF}{FB}=1$, and we are done.

Let the MacBeath inconic touch $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ at $D$, $E$, and $F$, respectively. By Brianchon on $AFBDCE$, lines $AD$, $BE$, and $CF$ concur.

Let $H_1,H_2,H_3$ be the reflections of $H$ over $BC,CA,AB$ respectively. Let $OH_1,OH_2,OH_3$ intersect $BC,CA,AB$ at $D,E,F$ respectively. We claim that these points $D,E,F$ are the desired points in the problem. Let $\angle A=\alpha,\angle B=\beta,\angle C=\gamma$. Note that $\angle BAH_1=90-\beta$, so $\angle BOH_1=180-2\beta$. Since $OB=OH_1$, $\angle DH_1B=\beta$. Similarly, $\angle DH_1C=\gamma.$ Additionally, note that $\triangle ABH_1$ and $\triangle ACH_1$ have the same circumradius, so $$\frac{BH_1}{CH_1}=\frac{\sin(90-\beta)}{\sin(90-\gamma)}.$$Then, by Ratio Lemma, $$\frac{BD}{CD}=\frac{BH_1}{CH_1}\cdot \frac{\sin \beta}{\sin\gamma}=\frac{\sin(90-\beta)\sin\beta}{\sin(90-\gamma)\sin\gamma}.$$Taking the cyclic product of this over $\alpha,\beta,\gamma$ yields 1 as everything cancels, so by Ceva's, $AD,BE,CF$ concur so we are done.

We look for an obvious placement for D,E,F, because this condition is not so convenient. First reflect H to obtain H_a, H_b, H_c. Then the condition is that H_aD+DO=..., so we make this a straight line by placing D as the intersection of H_aO and BC, and cyclically, so that the sum is equal to the circumradius. Now employ Ceva's to get the desired result. We can express these lengths as law of sines. We have <H_aBD=<H_aBC=<H_aAC=90-<C, and <DH_aB=<OH_aB=90-<H_aOB/2=90-H_aAB=<B. Then BD/DH_a (by law of sines) = sinB/cosC, and an analogous argument gives CD/DH_a=sinC/cosB. Dividing, we have BD/CD=sinBcosB/sinCcosC and cyclically. Then by Ceva's we're done $\blacksquare$

inellipse!

Note that by the existence of the MacBeath inconic, there is an ellipse with foci at $O$ and $H$ inscribed in $\triangle ABC$. Take $D, E, F$ to be the tangency points of the ellipse with sides $BC, AC, AB$, respectively. By Brianchon theorem, $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ concur. By ellipse properties $OD + DH = OE + EH = OF + FH$, so we are done. To be fair, this is more or less cheating. However, proving that the MacBeath inconic exists can be done easily using phantom point + angle chase, but I guess I am too lazy to write it up.

If $H_A,H_B,H_C$ are the reflections of $H$ about $BC,CA,AB$, $(D,E,F):=(OH_A\cap BC,OH_B\cap CA,OH_C\cap AB)$ satisfy the length condition with a constant sum of $R_{(ABC)}$. We prove concurrency by Ceva's: by the Law of Sines, \begin{align*} BD=\sin\angle BOH_A\cdot\tfrac{OD}{\sin\angle OBC}&=OD\cdot\tfrac{\cos2B}{\cos A}\\ CD=\sin\angle COH_A\cdot\tfrac{OD}{\sin\angle OCB}&=OD\cdot\tfrac{\cos2C}{\cos A}.\\ \end{align*}So $\tfrac{BD}{CD}=\tfrac{\cos2B}{\cos2C}$ and analogously $(\tfrac{CE}{AE},\tfrac{AF}{BF})=(\tfrac{\cos2C}{\cos2A},\tfrac{\cos2A}{\cos2B})$, whence multiplying gives $\tfrac{BD\cdot CE\cdot AF}{CD\cdot AE\cdot BF}=1$ as desired. $\square$

Let $A'$ be the reflection of $H$ across $BC$ and let $D$ be $OA'\cap BC$. Construct $E$ and $F$ similarly. Note that by orthocenter properties, we have that $A'\in (ABC)$, so we have that \[OD+DH=OD+DA'=OA'=R,\]and similarly, we also find that $OE+EH=OF+FH=R$ also. Now we must prove that the $AD$, $BE$, and $CF$ are concurrent. Through angle chasing, we get that \[\angle DHA'=\angle OAH=\angle BAC-\angle BAH-\angle OAC=a-(90-b)-(90-b)=180-(2b+a)=b-c,\]by orthocenter properties. Additionally, we also get that $\angle BCH=90-a$, which gives that \[\angle DHC=90-\angle DHA'-\angle DCH=c.\]Now using Law of Sines, we get that \[BD=DH*\frac{\sin b}{\cos c},\]and \[CD=DH*\frac{\sin c}{\cos b},\]which gives that \[\frac{BD}{CD}=\frac{\sin b\cos b}{\sin c\cos c}=\frac{2\sin b\cos b}{2\sin c\cos c}=\frac{\sin 2b}{\sin 2c}.\]Similarly, we get that $\frac{CE}{EA}=\frac{\sin 2c}{\sin 2a}$ and $\frac{AF}{FB}=\frac{\sin 2a}{\sin 2b}$. This gives us that \[\frac{BD}{CD}*\frac{CE}{EA}*\frac{AF}{FB}=1,\]which means that $AD$, $BE$, and $CF$ are concurrent by Ceva's Theorem, finishing the problem.

Note that $D$, $E$, and $F$ must be the tangency points of $\triangle ABC$ and the Macbeath ellipse, since the ellipse has foci $O$ and $H$. Then Brianchon's finishes.

We claim finding the smallest possible value of the quantity $OD+DH$ works. It's a classic result that this point $D$ is just $OH_A \cap BC$, where $H_A$ is the reflection of $H$ over $BC$, so \[OD+DH = OD+DH_A = R,\] which is obviously cyclically symmetric. Ceva's gives confirms the final requirement, as \[\prod \frac{BD}{DC} = \prod \frac{BH_A \cdot \sin \angle BH_AD}{CH_A \cdot \sin \angle CH_AD} = \prod \frac{\sin B \cos B}{\sin C \cos C} = 1. \quad \blacksquare\]

Let $(ABC) = \Gamma$. Let $AH \cap \Gamma = A_1$, $BH \cap \Gamma = B_1$ and $CH \cap \Gamma = C_1$. Obviously $A_1$ is a reflection of H over BC, because of famous orthocenter property, similarly for $B_1$ and $C_1$. Now let $OA_1 \cap BC = D$, $OB_1 \cap AC = E$ and $OC_1 \cap AB = F$. We will show that the D, E and F we defined satisfy the condition of the problem. Firstly $DO + DH = DO + DA_1 = OA_1 = R$, similarly $EO + EH = EO + EB_1 = OB_1 = R$ and $FO + FH = FO + FC_1 = OC_1 = R$ $\Rightarrow$ DO + DH = EO + EH = FO + FH = R. So it is only left to show that AD, BE and CF are concurrent. Now using the ratio lemma $\frac{AF}{FB} = \frac{AC_1}{BC_1} \cdot \frac{\sin \angle AC_1F}{\sin \angle BC_1F} = \frac{\sin \alpha \cdot \cos \alpha}{\sin \beta \cdot \cos \beta}$, similarly $\frac{BD}{DC} = \frac{BA_1}{CA_1} \cdot \frac{\sin \angle BA_1D}{\sin \angle CA_1D} = \frac{\sin \beta \cdot \cos \beta}{\sin \gamma \cdot \cos \gamma}$ and $\frac{CE}{EA} = \frac{CB_1}{EB_1} \cdot \frac{\sin \angle CD_1E}{\sin \angle AD_1E} = \frac{\sin \gamma \cdot \cos \gamma}{\sin \alpha\cdot \cos \alpha}$ $\Rightarrow$ by Ceva's $\frac{AF}{FB} \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = \frac{\sin \alpha \cdot \cos \alpha}{\sin \beta \cdot \cos \beta} \cdot \frac{\sin \beta \cdot \cos \beta}{\sin \gamma \cdot \cos \gamma} \cdot \frac{\sin \gamma \cdot \cos \gamma}{\sin \alpha\cdot \cos \alpha} = 1$ it follows that AD, BE and CF are concurrent, which was what we needed to show $\Rightarrow$ the D, E and F we choose work and we are ready.

Pretty interesting problem. Actually the proof of the problem generalizes for an arbitrary pair of isogonal conjugates $P$ and $Q$ in $\triangle ABC$. We define by $H_a$ , $H_b$ and $H_c$ the reflections of the orthocenter $H$ across sides $BC$ , $AC$ and $AB$ of $\triangle ABC$. Then, let $D$ , $E$ and $F$ be the intersections of $\overline{OH_a}$ , $\overline{OH_b}$ and $\overline{OH_c}$ with sides $BC$ , $AC$ and $AB$ respectively. The following is the pith of the problem. Claim : Points $D$ , $E$ and $F$ lie on an inellipse $\mathcal E$ of $\triangle ABC$ with foci $O$ and $H$. Proof : First note that since $H_a$ is the reflection of the orthocenter $H$ across side $BC$ , $DH_a=DH$. Thus, \[OD+DH = OD + DH_a = R \]where $R$ is the circumradius of $\triangle ABC$. Similarly we can compute that, \[OD+DH = OE + EH = OF + FH = R\]which implies that $D$ , $E$ and $F$ lie on an ellipse $\mathcal E$ with foci $O$ and $H$. Next, note that due to the reflection, \[\measuredangle ODC = \measuredangle H_a DB = \measuredangle BDH\]which implies that side $BC$ is in fact tangent to $\mathcal E$ at $D$. A similar argument shows that $\mathcal E$ is also tangent to $\triangle ABC$ on sides $AB$ and $AC$ at points $F$ and $E$ respectively, which proves the claim. We finish by simply noting that Brianchon's Theorem on $AFBDCE$ implies that $AD, BE$ and $CF$ concur, as desired.

???????????? Since $O$ and $H$ are isogonal conjugates, there is an ellipse with them as foci that is tangent to all three sides of the triangle. Then Brianchon finishes.