Solution

Synthetic Solution Let $X = I_AT \cap (ABC)$ and let $M$ be the midpoint of interior arc $BC$ It is widely known that $AT, AD$ are isogonal lines and $ID \parallel MX$ ($ID \parallel MX$ can be proved easily by angle chasing using that $\triangle ATI_A \sim \triangle AID$) It is sufficient to prove that $\angle TXM + \angle DIK = 90^{\circ}$ Here, $\angle TXM = \angle TAM = \angle MAD$ and $\angle MAD + \angle DIK = 90^{\circ}$ Let $I'$ be the antipode of $I$ wrt $(AID)$, and $\angle MAD = \angle IAD = \angle II'D = 90^{\circ} - \angle DII'$ So, $I_AT \perp KI$

After Root BC inversion centered at A. I ---> I_A D ----> T (Due to property of mixtilinear circle) Just observe similarity and angle chase We are done

Let $X=\overline{KI} \cap \overline{I_AT}$ and $M$ be midpoint of $\overline{AI}$.We will prove $M,K,I_A,X$ cyclic Take $\sqrt{bc}$ inversion. \[ I \longleftrightarrow I_A, \quad T \longleftrightarrow D \]$$\angle MKI =\angle ADI = \angle AI_AT$$hence we get $\overline{KI} \perp \overline{I_AT}$

Obviously we have to prove that $\angle AIK + \angle TI_AI =90$. $K$ is the center of $(AID)$, hence $\angle AIK = 90 - \angle IDA$, so it remains to prove that $\angle IDA = \angle TI_AI$. Let $I'$ be the symmetric point of $I$ with respect to $T$, $N$ the midpoint of arc $BAC$ in $(ABC)$ and $M$ be the midpoint of arc $BC$ in $(ABC)$ which does not contain $A$. Then we have that $TI'=TI$ beacuse of symmetry and $\angle MTI = \angle MTN = 90$, hence $MI'=MI=MI_A$. From this equality we conclude that $\angle II'I_A=90$ and $NAI'I_A$ is cyclic. Now let $X$ be a point on $BC$ such that $AX \perp BC$. It is well known that $AT, AD$ are isogonal and hence we get that $\angle BAD = \angle TAC$ and $\angle ABD = \angle ABC= \angle ATC$ (from the cyclic $(ABTC)$), so triangles $ABD$ and $ATC$ are similar. From this similarity and the cyclic $NAI'I_A$ we get that $\angle ADX = \angle ADB = \angle ACT = \angle ANT = \angle II_AI'$ From the above equality we get that the right angle triangles $AXD$ and $II'I_A$ are similar. In these similar triangles $I_AT$ and $DI$ are medians of the respective sides (it is well known $DI$ bisects $AX$). Hence, because of similarity, we get that $\angle IDA = \angle TI_AI$, as needed.

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