Why is it true that the angle $\angle AIC = 90^{\circ}+ \frac{\angle ABC}{2}$ Consider the following picture (borrowed from the web). It is a well-known fact that I most recently saw on page 12 of Coexeter's *Introduction to Geometry that the angle $\angle AIC = \frac{\angle ABC}{2} + 90^{\circ}$.
I am having trouble seeing this and was wondering if someone could either show me why or direct me to a reference(it's very likely this is a repeated question)

 A: It's immediate by considering the right triangles:

A: There are $6$ pairwise equal angles that meet at the point $I$. Let's call them $\alpha$, $\beta$, $\gamma$ (against $A$, $B$, $C$ respectively). Then $\alpha+\beta+\gamma=180^\circ$. At the same time, we have from the triangle $IM_cB$ that $\beta+90^\circ+\frac{\angle ABC}{2}=180^\circ$. It gives the desired relation for $\angle AIC$, which is $\alpha+\gamma$.
A: we have $$\angle{ABC}=180^{\circ}-\frac{\alpha}{2}-\frac{\beta}{2}=\alpha+\beta+\gamma-\frac{\alpha+\gamma}{2}=\frac{\alpha+\gamma}{2}+\beta$$
and $$\frac{\alpha+\gamma}{2}+\beta=\frac{\beta}{2}+90^{\circ}$$ since $$\alpha+\beta+\gamma=180^{\circ}$$
A: From
$$\widehat{AIC}+\widehat{IAC}+\widehat{ICA}=180^{\circ}$$
$$\widehat{AIC}=\widehat{ABI}+\widehat{BAI}+\widehat{IBC}+\widehat{BCI}=\widehat{ABC}+\widehat{BAI}+\widehat{BCI}$$
Also $\widehat{IAC}=\widehat{BAI}$ and $\widehat{ICA}=\widehat{BCI}$.
Altogether
$$\widehat{AIC}=180^{\circ}-\widehat{IAC}-\widehat{ICA}=180^{\circ}-\left(\widehat{AIC}-\widehat{ABC}\right)$$
$$2\cdot \widehat{AIC}=180^{\circ}+\widehat{ABC} \Rightarrow \widehat{AIC}=90^{\circ}+\frac{\widehat{ABC}}{2}$$
