So many more things that can be found in this diagram than this result. Its annoying how it must reduce to some ratios: Let us re-phrase it: $ABC$ be a triangle. $DEF$ be the orthic triangle. $A'B'C'$ be the intouch triangle. Suppose $H \in B'C'$, prove that if $I$ is incentre we have $IB', IC'$ meet opposite sides of tangencies of the A-excircle. Let $X$ be the tangency with the A-excircle with $AB$. Note, by the converse of brianchon theorem, there exists a circle tangent to $BFEC$, which is of course the incircle. Now, let $k$ denote the ratio $\dfrac{AE}{AB} = \dfrac{AF}{AC} = \dfrac{EF}{BC}$. By Pithots, we have $EF + BC = BC + kBC = FB + EC = AC + AB - k(AC + AB)$ or $k(AB+BC+AC) = AC + AB - BC$ or $k = \dfrac{s-a}{s}$ using usual notation. Since $k = \dfrac{AE}{AB} = \dfrac{s-a}{s} = \dfrac{AB'}{AX}$ we have $BE \parallel B'X$ and so we are done.

Let $I$ be the incenter of $\triangle ABC$,$R,r$ be the radii of its circumcircle and incircle respectively,and denote by $B_1$ the pedal of $I$ on $AC$,then $A'C_1\perp AB\iff A',I,C_1$ are collinear $\iff \angle A'IA_1=\angle ABC\iff \tan \angle A'IA_1=\tan \angle ABC$ $\iff \frac{A'A_1}{r}=\frac{AC}{r}=\frac{AC}{BH}\iff BH=r\iff \cos \angle ABC=\frac{BH}{2R}=\frac{r}{2R}$ $\iff \cos \angle BAC+\cos \angle ACB=$ $1+\frac{r}{R}-\cos \angle ABC=1+\frac{r}{2R}=1+\cos \angle ABC=2\cos^2 \frac{\angle ABC}{2}=$ $\cot \frac{\angle ABC}{2}\sin \angle ABC$ $\iff \frac{\sin \angle HBA_1}{BC_1}+\frac{\sin \angle HBC_1}{BA_1}=\frac{\cos \angle ACB}{BC_1}+\frac{\cos \angle BAC}{BA_1}=\frac{\sin \angle ABC}{r}$ $\iff A_1,C_1,H$ are collinear, and the conclusion follows,as desired. $Q.E.D.$

$\iff \frac{\sin \angle HBA_1}{BC_1}+\frac{\sin \angle HBC_1}{BA_1}=\frac{\cos \angle ACB}{BC_1}+\frac{\cos \angle BAC}{BA_1}=\frac{\sin \angle ABC}{r}$ $\iff A_1,C_1,H$ are collinear, Arab, can you explain it clearly, i don't understand.

Lemma $D$ is a point on segment $BC$ of $\triangle ABC$,then $\frac{\sin \angle CAD}{AB}+\frac{\sin \angle BAD}{AC}=\frac{\sin \angle BAC}{AD}$. Proof Denote by $E$ the second point of intersection of $AD$ and the circumcircle of $\triangle ABC$ and let $R$ be the circumradius of $\triangle ABC$,then we obtain that \begin{align*}\frac{\sin \angle CAD}{AB}+\frac{\sin \angle BAD}{AC}& =(\frac{CE}{AB}+\frac{BE}{AC})\cdot \frac{1}{2R}\\& =\frac{CE\cdot AC+BE\cdot AB}{AB\cdot AC}\cdot \frac{1}{2R}\\& =\frac{[ABEC]}{[ABC]}\cdot \frac{\sin \angle BAC}{\sin \angle ABE}\cdot \frac{1}{2R}\\& =\frac{AE}{AD}\cdot \frac{\sin \angle BAC}{AE}\\& =\frac{\sin \angle BAC}{AD}\end{align*} In the original problem,let $H'=BH\cap A_1C_1$ and apply the lemma,then we get that $H'\equiv H$.

Arab, thank you very much.

Arab wrote: Lemma $D$ is a point on segment $BC$ of $\triangle ABC$,then $\frac{\sin \angle CAD}{AB}+\frac{\sin \angle BAD}{AC}=\frac{\sin \angle BAC}{AD}$. sorry Arab but the proof of the lemma is trivial and non-constructional.and btw it is subtended angle theorem. proof we just equalize the area. $\frac{1}{2} \angle BAC. AB.AC=\frac{1}{2} \angle BAD .AD.AB+\frac{1}{2} \angle CAD. AD.AC$. so dividing both sides with $AD.AC.AB$ .the lemma is proved.btw my solution is same as Arab as well.

Can someone explain how by the converse of Brianchon's theorem, there exists a circle tangent to $BFEC$ in IDMasterz solution. I thought Brianchon's theorem only applies to hexagons.

We have AC' = CA1 and AC1 = CA'. Let altitude from C to AB meet A1C1 at Y and altitude from A to BC meet A1C1 at X. By thales we have : C1Y/YA1 = A'C/CA1 = CA'/AC1 = C1X/XA1 ---> X and Y are same so orthocenter of triangle ABC lies on A1C1. The other part has same approach.