Prove that the intouch point of the medial triangle is collinear with a vertex and an extouch point of a reference triangle Let $E$ be the midpoint of $AB$, $F$ be the midpoint of $AC$. I need to prove that the point at which the Spieker circle (the incircle of the medial triangle) touches $EF$ is on the line $AT_a$, where $T_a$ is the extouch point of the triangle on $BC$. I have tried homothety and similar triangles, both to no avail. Any solutions would be much appreciated. Thanks

First of all notice that $\triangle AEF \cong \triangle DFA$ (in the same order of vertices), where $D$ is the midpoint of $BC$. Now let the incenter of $\triangle AEF$ touch $EF$ at $H$. Now by the symmetry of the abovementioned triangles we have that $H$ and $X$ are reflections of each other over the midpoint of $EF$. In otherwords $X$ is the touching point of the excircle of $\triangle AEF$ opposite the vertex $A$. (It's a well-known property that the touching point of the incircle and the corresponding excircle are reflections of each other over the midpoint of the corresponding side).

Now take the excirle of $\triangle AEF$ opposite the vertex $A$ and by homothety send it to the excircle of $\triangle ABC$ opposite the vertex $A$. Now using the fact that $EF \parallel BC$ is not hard to conclude that the touching points are collinear with the center of homothety, i.e. $A,X,T_a$ are collinear. Hence the proof.