Prove that $CH \bot AB$

[Tahmid Hasan]

Let $(O),(I)$ be the circumcircle and incircle of $\triangle ABC$ respectively.$BC,CA,AB$ are tangent to $(I)$ at $M,N,P$ respectively. $E,F$ are midpoints of $PN,PM$ rspectively. $CO$ meets $(I)$ at point $D$. $DE,DF$ meet $(I)$ at $K,L$ respectively, $H=AK \cap BL$. Prove that $CH \bot AB$.

[*Mahi*]

The problem seemed quite nice to me.
My solution steps:

1. Proving $\angle KAB = \angle DAC$ and $\angle LBA = \angle DBC$ using symmetry.
2. Using trig cheva to prove $\angle HCB = \angle DCA = \angle OCA$, which also implies $CH \perp AB$.

($(O)$ is quite irrelevant here and just used to prove the last part, the main fact here is $H$ is the isogonal conjugate of $D$, similar facts of which is used to solve many problems.

Just noticed, this is my 900th post a big congratulation to myself

[Nadim Ul Abrar]

$D,H$ are isogonal Conjugates

[Tahmid Hasan]

The problem seemed quite nice to me.
My solution steps:

1. Proving $\angle KAB = \angle DAC$ and $\angle LBA = \angle DBC$ using symmetry.
2. Using trig cheva to prove $\angle HCB = \angle DCA = \angle OCA$, which also implies $CH \perp AB$.

($(O)$ is quite irrelevant here and just used to prove the last part, the main fact here is $H$ is the isogonal conjugate of $D$, similar facts of which is used to solve many problems.

Would you kindly elaborate?
My solution:$DE.KE=PE.NE=AE.IE \Rightarrow AKID$ is cyclic.
Since $I$ lies on the perpendicular bisetor of $KD, AI$ bisects $\angle AKD$.
So $AH$ is the isogonal of $AD$ in $\triangle ABC$. Similarly $BH$ is the isogonal of $BD$ in $\triangle ABC$.
Hence $H$ is the isogonal conjugate of $D$ wrt $\triangle ABC \Rightarrow CH$ is the isogonal of $CO \Rightarrow CH \bot AB$.
Note: Actually $(I),CO$ have two intersection points but both can be dealt with the same reasoning.

[*Mahi*]

Lines $KA$ and $DA$ ar symmetric WRT $AI$, (Do I need to elaborate more here? This can be proved using pole-poler and some little other things, and I left that as an exercise), which implies $\angle KAB = \angle DAC$.