Estonian Mathematical Olympiad

As the $A$-bisector is also the altitude, we have $BD\perp AD$. Let $I$ be the incenter of $ABC$ and $E^{\prime }\ne B$ the intersection of $BI$ and the circumcircle of $ABD$ (Fig. 2). As the angles $\angle AB{E}^{\prime }=\angle DB{E}^{\prime }$ correspond to the arcs $E^{\prime }A$ and $E^{\prime }D$ of this circle, we have $E^{\prime }A=E^{\prime }D$. Thus $E^{\prime }$ lies on the perpendicular bisector of $AD$, meaning $E^{\prime }=E$. Hence $ABDE$ is a cyclic quadrilateral (Fig. 3). It remains to verify that $\angle EDI+\angle CID=90^{\circ }$, which would yield $CI\perp DE$, as desired. To show this, let $\angle ACB=\angle ABC=\gamma$. Then $$\angle EDI=\angle EDA=\angle EBA=\frac{\gamma }{2},$$ whereas $$\angle CID=180^{\circ }-\angle CDI-\angle DCI=180^{\circ }-90^{\circ }-\frac{\gamma }{2}=90^{\circ }-\frac{\gamma }{2}.$$ Thus $\angle EDI+\angle CID=90^{\circ }$, finishing the proof.

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Alternative solution.

As the $A$-bisector is also the altitude, we have $BD\perp AD$. Thus the perpendicular bisector of $AD$ is parallel to $BC$, therefore containing the midline of $ABC$ and bisecting $AB$ and $AC$. Let $L$ be the midpoint of $AB$ and $I$ the incenter of $ABC$ (Fig. 4). Then $\angle LEB=\angle CBE=\angle LBE$, meaning that $LBE$ is isosceles and $LE=LB=LA$. Thus $L$ is the circumcenter of $ABE$ and $AB$ is a diameter of the circle. Now as $ADB=90^{\circ }$, the point $D$ also lies on this circle, meaning $ABDE$ is cyclic. We proceed like in Solution 1.

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Alternative solution.

As the $A$-bisector is also the altitude, we have $BD\perp AD$. Thus the perpendicular bisector of $AD$ is parallel to $BC$, therefore containing the midline of $ABC$ and bisecting $AB$ and $AC$. Let $L$ and $M$ be the midpoints of $AB$ and $AC$, respectively. Then $\angle LEB=\angle CBE=\angle LBE$, meaning that $LBE$ is isosceles and $LE=LB=LA$. Also $MC=MA=LB=LE$, and as $LM$ is the midline of $ABC$, we have $LM=\frac{1}{2}BC=CD$. Let $N$ be the intersection of $DE$ and $AC$; depending on the orientation of points (Figures 5 and 6) we have $$\begin{gathered} CN=CM\pm MN, \\ CD=LM=LE\pm EM=CM\pm ME. \end{gathered}$$ Triangles $NCD$ and $NME$ are similar by parallel sides, giving $\frac{CD}{ME}=\frac{CN}{MN}$. Substituting in the previous equation, we get $\frac{CM\pm ME}{ME}=\frac{CM\pm MN}{MN}$, from which $\frac{CM}{ME}\pm 1=\frac{CM}{MN}\pm 1$, which in turn implies $\frac{CM}{ME}=\frac{CM}{MN}$. Therefore $ME=MN$, from which $CD=CN$. The claim of the problem follows, as the $C$-bisector in the isosceles triangle $CDN$ and the side $DN$ are perpendicular.

Fig. 5 Fig. 6

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Alternative solution.

Let $I$ be the incenter of $ABC$. Clearly $D$ is the point of tangency of the incircle and side $BC$. Let $N$ be the tangency point with side $AC$ (Fig. 7), then $CD=CN$. In the isosceles triangle $CDN$, the line $CI$ is both the angle bisector and altitude, so $DN$ and $CI$ are perpendicular. It remains to show that $E$ lies on $DN$. By the Iran lemma, the angle bisector $BI$ and the extensions of the incircle chord $DN$ and the midline corresponding to $BC$ intersect at one point. Of those, the first and the third intersect at $E$, so $E$ must lie on $DN$, as desired.

Fig. 7