$\angle AFI =90$

Question

Let triangle $ABC$ inscribed $(O)$ and $(I)$ inscribed triangle $ABC$. $AH,ID$ are perpendicular to $BC$. $AI$ cuts $(O)$ at $E$. $DE$ cuts $(O)$ at $F$. $BC$ cuts $AF$ at $K$.

a) Prove that $A,I,K,H$ lie on one circle.

b) $EH$ cuts $(O)$ at $L$. $FL$ cuts $BC$ at $J$. Prove that tangent of $(O)$ from $F$ passes mid-point of $JK$

Here is what I try:

Because $A,I,K,H$ lie on one circle so I want to prove that $FIDK$ is cyclic.

So it lead to prove $\angle AFI =90$ which is my problem.

I need this a lots for my next day, thanks you for all your helps.

Answer 1

Comment , A, I,K,H are cyclic:

Draw circle $d_1$ passing point A, K and H, since $\angle AHK=90^o$ then AK is the diameter of the circle $d_1$. Now we have to show $\angle AIK=90^o$. For this we connect E to K and joint midpoint N of AK to mid point M of KE, we have $MN||AE$. KI is common chord of two circles $d_1$ and $c_1$, so $MN\bot KI$ which results in $AE\bot KI$ which means $\angle AIK=90^o$

Answer 2

Let $M,N$ be the two other tangent points to the incircle of triangle $ABC$. Draw the circle with diameter $AI$, call it $\omega$. Notice that $\omega$ passes through $M,N$ because $\angle AMI = \angle ANI = 90^{\circ}$.

We show that $F$ lies on $\omega$ and that should imply $\angle AFI=90^\circ$ as you want.

Connect $FB, FC, FM, FN$. Notice since $FD$ bisects $\angle BFC$, we have ${FB\over BD} = {FC\over CD}$. Also since $BM=BD$ and $CN=CD$ we have

$${FB\over BM} = {FC\over CN} \tag{1}$$

Also since $\angle FBA = \angle FCA$ we have

$$\angle FBM = \angle FCN \tag{2}$$

Equations (1) and (2) imply $\triangle FBM$ and $\triangle FCN$ are similar. The following angle chasing leads to,

$$\angle BFM = \angle CFN \implies \angle MFN = \angle BFC = \angle BAC = \angle MAN$$

And this implies $A,F,M,$ and $N$ are concyclic, which means point $F$ lies on $\omega$.

$\underline{\text{Disclaimer}}:\enspace$ this idea of proving does not originate from myself, but from the author of this question (on the OP's partial work):Prove that $P=RA'\cap EF$, then $DP\perp EF$.