If $D$ is tangent point of $BC$ with incircle of $ABC$ and $T$ is intersection of $A$-mixtilinear inclrcle with $(ABC)$, then $\angle BTA=\angle CTD$.

Question

Problem: Let $ABC$ be a triangle and let its incircle touch $BC$ at $D$. Let $T$ be the tangency point of the $A$-mixtilinear incircle with circumcircle of $ABC$. Prove that $∠BTA = ∠CTD$.

I recognised that it suffices to show that $KT$ is isogonal to $DT$.Then the first thing that came to my mind is harmonic quadrilateral.I don't whether it is true or not.I am new to the topic of isogonals. Help me how to start??

Edit:I asked to prove that the above marked angles are equal.The question which marked as possible duplicate is not the same.Angles which were to be proven are different.I hope this convince you this as different question.I use mobile so geogebra in android is very difficult so I drew diagram manually.