IRL_ABooklet_2023

Extend $AD$ and $AE$ to meet the circumcircle of $△ABC$ at $F$ and $G$, respectively. Let $O$ be the intersection point of $BG$ and $CF$.



Since $\angle BAD=\angle CAE$, we have $$\angle BCO=\angle OGF=\angle BAD=\angle CAE=\angle CBO=\angle OFG,$$ subtending equal arcs. Hence $OB$ and $OC$ are tangents to the circumcircles of $△BAD$ and $△CAE$ by the Alternate Segment Theorem, and $△OBC$ is isosceles with $|OB|=|OC|$. Then $O$ must be on the common tangent to the circumcircles of $△BAD$ and $△CAE$, since it has the same power with respect to them. Since $OA$ is tangent to the two circles, $|OA|=|OB|=|OC|$ hence $O$ is the circumcentre of $△ABC$, hence $BG$ and $CF$ are diameters, thus $\angle BAE=\angle CAD=90^{\circ }$.

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Alternative solution.

Let $M,N$ be the centres of the circumcircles of triangles $BAD$ and $CAE$, respectively. These two circles are tangent at $A$ iff $M,A,N$ are collinear. We define $\alpha =\angle BAD=\angle CAE$, $\beta =\angle MAB=\angle MBA$, and $\gamma =\angle NAC=\angle NCA$.



We then have $\angle BMA=180^{\circ }-2\beta$ and $\angle ANC=180^{\circ }-2\gamma$. If $M$ and $D$ are on opposite sides of $AB$, we have $\angle BDA=180^{\circ }-\frac{1}{2}\angle BMA=90^{\circ }+\beta$ and $\angle ADE=180^{\circ }-\angle BDA=90^{\circ }-\beta$. Similarly, $\angle AED=90^{\circ }-\gamma$. We now obtain $$\angle DAE=180^{\circ }-\angle ADE-\angle AED=\beta +\gamma .$$ Because $M,A,N$ are collinear, this implies $2\alpha +2\beta +2\gamma =180^{\circ }$, hence $\alpha +\beta +\gamma =90^{\circ }$. Finally, $\angle BAE=\angle BAD+\angle DAE=\alpha +\beta +\gamma =90^{\circ }$ and $\angle DAC=\angle DAE+\angle EAC=\beta +\gamma +\alpha =90^{\circ }$.

If $M$ and $D$ were on the same side of $AB$, the above argument, with $\beta$ replaced by $-\beta$ in the formulas for $\angle BDA$, $\angle ADE$ etc., gives the same result. We then see that $\angle BDA=90^{\circ }+\alpha >90^{\circ }$, which implies that $M$ and $D$ must actually be on opposite sides of $AB$.