An equivalence relation in a triangle

Prove that in any triangle $ABC$ the following equivalence relation holds:

$a^2=ab+c^2 \Leftrightarrow \hat{A}= 90^\circ + \frac{\hat{C}}{2}$

Solution

We are working on the following shape.

Let $I$ be the incenter of the triangle. Thus,

$\widehat A = {90^0} + \frac{{\widehat C}}{2} \Leftrightarrow \widehat A = \widehat{AIB} \Leftrightarrow \widehat{BAI} = \widehat{AEB} = \frac{\hat{A}}{2}$

meaning that $AB$ is tangent to the circumcircle of the triangle $AIE$ . Thus,

(1) $\begin{equation*}c^2 = BI \cdot BE \end{equation*}$

Hence,

$\begin{align*} \frac{BI}{BE} = \frac{a+c}{a+b+c} &\overset{(1)}{\Leftrightarrow } c^2 = \frac{a+c}{a+b+c} \cdot BE^2 = \frac{a+c}{a+b+c} \left ( ac - \frac{b^2}{\left ( a+c \right )^2} \right ) \\ &\Leftrightarrow c^2 = \frac{ac\left ( a+c-b \right )}{a+c} \\ &\Leftrightarrow a^2 = ab + c^2 \end{align*}$