Saudi Arabian Mathematical Competitions

It is well known that the bisector of the angle at vertex $A$ and the perpendicular bisector of side $BC$ intersect on the circumcircle of triangle $ABC$. We give here two solutions.

Solution 1. We have
$$\begin{aligned} \measuredangle JDI & = 180^\circ - \measuredangle EDJ = 180^\circ - \frac{1}{2} \measuredangle ADB = 180^\circ - \frac{1}{2}(\measuredangle DAC + \measuredangle ACB) \\ & = 180^\circ - \left(\frac{1}{4} \measuredangle BAC + \frac{1}{2} \measuredangle ACB\right) \end{aligned}$$
On the other hand
$\measuredangle ICJ = \frac{1}{2} \measuredangle ACB + \frac{1}{2} \measuredangle BCE = \frac{1}{2} \measuredangle ACB + \frac{1}{2} \measuredangle BAE = \frac{1}{2} \measuredangle ACB + \frac{1}{4} \measuredangle BAC$.
Therefore, quadrilateral $C I D J$ is cyclic.

We deduce that
$$\measuredangle JIC = \measuredangle JDC = \measuredangle EDJ = \measuredangle ICJ,$$
and triangle $CIJ$ is isosceles.

Solution 2. It is well known that $EC = EI$. This is because
$$\measuredangle ICE = \measuredangle ICD + \measuredangle DCE = \measuredangle ACI + \measuredangle BAE = \measuredangle ACI + \measuredangle IAC = \measuredangle EIC.$$

Therefore, the bisector of angle $\angle CEI$ is the perpendicular bisector of segment $CI$. But $J$ is on this bisector. Hence, $JC = JI$ and triangle $CIJ$ is isosceles.