SAUDI ARABIAN MATHEMATICAL COMPETITIONS

First, we will show that $D$ is the orthocenter of triangle $ABC$. Denote $A^{\prime }=AD\cap BC$, $B^{\prime }=BD\cap CA$, $C^{\prime }=CD\cap AB$. Since $\angle DAB=\angle DCB$, we have $AC{A}^{\prime }{C}^{\prime }$ is cyclic. Similarly, $AB{A}^{\prime }{B}^{\prime }$ is also cyclic. So we have $$\angle D{A}^{\prime }B=\angle D{B}^{\prime }A,\angle D{A}^{\prime }C=\angle D{C}^{\prime }A\text{ and }DA\cdot D{A}^{\prime }=DB\cdot D{B}^{\prime }=DC\cdot D{C}^{\prime },$$ so $BC{B}^{\prime }{C}^{\prime }$ is also cyclic. Thus $\angle D{B}^{\prime }A=\angle D{C}^{\prime }A$ which implies that $\angle D{A}^{\prime }B=\angle D{A}^{\prime }C$, but $\angle D{A}^{\prime }B+\angle D{A}^{\prime }C=180^{\circ }\to A{A}^{\prime }\perp BC$. It leads to $B{B}^{\prime }\perp CA$, $C{C}^{\prime }\perp AB$ then $D$ is the orthocenter of triangle $ABC$.

Continue, denote $M$ as the intersection of $AE$ and $BC$ then $\angle EBM=\angle EAB$ implies that $MB$ is tangent to $(ABE)$. Similarly, $MB$ is tangent to $(ACE)$ then $$ME\cdot MA=M{B}^2=M{C}^2=M{B}^{\prime 2}=M{C}^2$$ (since $MB=MC=M{B}^{\prime }=M{C}^{\prime }$). So $M{B}^{\prime }$ is tangent to $(AE{B}^{\prime })$, thus $\angle AE{B}^{\prime }=\angle M{B}^{\prime }C=\angle C$. Similarly, $\angle AE{C}^{\prime }=\angle B$ then $$\angle B^{\prime }E{C}^{\prime }=\angle AE{B}^{\prime }+\angle AE{C}^{\prime }=\angle B+\angle C=180^{\circ }-\angle A.$$ Thus $A{B}^{\prime }E{C}^{\prime }$ is cyclic, but $A{B}^{\prime }D{C}^{\prime }$ is cyclic, so five points $A,B^{\prime },C^{\prime },D,E$ are concyclic, which implies that $$\angle AED=\angle A{B}^{\prime }D=\angle A{C}^{\prime }D=90^{\circ }.$$ Therefore, $ADE$ is a right triangle. $\square$