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Let triangle $ABC$ with $(I)$ is incircle and $(I)$ tangent to $BC,CA,AB$ at $D,E,F$, respectively. Let $H\in EF$ such that $DH\perp EF$. Prove that $H$ and othorcenters of $\Delta AEF$ and $\Delta ABC$ are colinear.

Here are what i have done:

Let $(O)$ is the circumcircle of $\Delta ABC$ and $AK$ is the diameter of this circle. I have proved that $H,I,K$ are colinear and I wonder if it is useful.

Let $K=EF\cap BC$ then $(K,D,B,C$ are harmonic range adn thus $HD$ is bisector of $\widehat{BHC}$. Somebody can help me!
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Dat Tran
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  • It is fairly doable with analytic geometry. – user10354138 Sep 14 '21 at 23:24
  • can you give me some hint? Thank you – Dat Tran Sep 15 '21 at 00:17
  • Put the incentre $I$ at origin, and let $D=(0,-1)$. Then $E=(\sin C,\cos C)$, $F=(-\sin B,\cos B)$ and so the orthocentre of $\triangle AEF$ is $H_{AEF}=(\sin C-\sin B,\cos B+\cos C)$. Similarly $x_H=\frac{\sin C-\sin B}2+\sin(\frac{B-C}2)\cos(\frac{B-C}{2})$ and $y_H=\frac{\cos B+\cos C}2-\sin^2(\frac{B-C}2)$. Now compute $x_B,x_C$ and hence the coordinates of $H_{ABC}$. Then the problem reduces to verifying a trig identity. – user10354138 Sep 15 '21 at 04:42
  • Why can you find orthorcenter of $AEF$ while you don't have the coordinate of $A$. – Dat Tran Sep 15 '21 at 09:04
  • Well-known construction: The reflection of the orthocentre across a side of the triangle lies on the circumcircle (Proof: angle-chase). Now $\triangle AEF$ is isosceles so $AI$ is the perpedicular bisector of $EF$, and since $\angle AEI=90^\circ$ $AI$ is also the circumdiameter of $\triangle AEF$. So you can just reflect $I$ in $EF$ to get the orthocentre of $\triangle AEF$. – user10354138 Sep 15 '21 at 09:28

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