Construction and Properties of Mixtilinear Incircles

Mixtilinear incircle is a circle tangent to two sides of a triangle and to the triangle's circumcircle. In every triangle there are three mixtilinear incircles, one for each vertex. The diagram below shows the mixtilinear incircle corresponding to vertex $B$ in $\Delta ABC:$

Consider an inversion in the circle $C(B,\sqrt{ac}).$

Let a generic point $X$ be mapped on point $X'.$ Since $BA\cdot BA'=ac,$ $BA'=c;$ similarly, $BC'=a.$ It follows that $\Delta ABC=\Delta A'BC'.$ The circumcircle $(ABC)$ maps onto a straight line passing through both $A'$ and $C'$ such that it maps onto $A'C'.$ Thus the image of the mixtilinear circle is tangent to the three lines $BA$ (i.e., $BA'$), $BC$ (i.e., $BC'$), and $A'C',$ and, is, therefore, the excircle of $\Delta A'BC'.$ Because of the symmetry in the angle bisector $BI,$ it is also tangent to $AC$ and so is the excircle of $\Delta ABC.$

Now, it is easy to see that $BF'=p,$ making $BF=\displaystyle\frac{ac}{p}$ and, thus proving the first part.

For the second part, recollect that $BI^{2}=\displaystyle\frac{(p-b)ac}{p},$ whereas $BD=p-b,$ where $D$ is the point of tangency of the incircle $(I)$ with side $BC.$ We then have

$\displaystyle\frac{BI^2}{BF^2}=\frac{(p-b)ac}{p}\cdot\frac{p^2}{(ac)^2}=\frac{p(p-b)}{ac}.$

On the other hand,

$\displaystyle\frac{BD^2}{BI^2}=\frac{(p-b)^{2}p}{(p-b)ac}=\frac{p(p-b)}{ac}.$

This shows that $BI^{2}=BD\cdot BF,$ making $\Delta BIF$ right (at $I.)$

Note that this property explains the construction of mixtilinear circles discussed elsewhere.

I was pointed to the proof of the first relation at the community blog of the artofproblemsolving site by Emmanuel Antonio José García.