International Mathematical Olympiad

Throughout the solution, we refer to $\angle A,\angle B,\angle C,\angle D$, and $\angle E$ as internal angles of the pentagon $ABCDE$. Let the perpendicular bisectors of $AC$ and $BD$, which pass respectively through $B$ and $C$, meet at point $I$. Then $BD\perp CI$ and, similarly, $AC\perp BI$. Hence $AC$ and $BD$ meet at the orthocenter $H$ of the triangle $BIC$, and $IH\perp BC$. It remains to prove that $E$ lies on the line $IH$ or, equivalently, $EI\perp BC$.

Lines $IB$ and $IC$ bisect $\angle B$ and $\angle C$, respectively. Since $IA=IC,IB=ID$, and $AB=BC=CD$, the triangles $IAB,ICB$ and $ICD$ are congruent. Hence $\angle IAB=\angle ICB=\angle C/2=\angle A/2$, so the line $IA$ bisects $\angle A$. Similarly, the line $ID$ bisects $\angle D$. Finally, the line $IE$ bisects $\angle E$ because $I$ lies on all the other four internal bisectors of the angles of the pentagon.

The sum of the internal angles in a pentagon is $540^{\circ }$, so $$\angle E=540^{\circ }-2\angle A+2\angle B.$$ In quadrilateral $ABIE$, $$\begin{array}{rl}\angle BIE& =360^{\circ }-\angle EAB-\angle ABI-\angle AEI=360^{\circ }-\angle A-\frac{1}{2}\angle B-\frac{1}{2}\angle E\\ & =360^{\circ }-\angle A-\frac{1}{2}\angle B-\left(270^{\circ }-\angle A-\angle B\right)\\ & =90^{\circ }+\frac{1}{2}\angle B=90^{\circ }+\angle IBC,\end{array}$$ which means that $EI\perp BC$, completing the proof.

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Alternative solution.

We present another proof of the fact that $E$ lies on line $IH$. Since all five internal bisectors of $ABCDE$ meet at $I$, this pentagon has an inscribed circle with center $I$. Let this circle touch side $BC$ at $T$.

Applying Brianchon's theorem to the (degenerate) hexagon $ABTCDE$ we conclude that $AC,BD$ and $ET$ are concurrent, so point $E$ also lies on line $IHT$, completing the proof.

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Alternative solution.

We present yet another proof that $EI\perp BC$. In pentagon $ABCDE$, $\angle E<180^{\circ }⟺\angle A+\angle B+\angle C+\angle D>360^{\circ }$. Then $\angle A+\angle B=\angle C+\angle D>180^{\circ }$, so rays $EA$ and $CB$ meet at a point $P$, and rays $BC$ and $ED$ meet at a point $Q$. Now, $$\angle PBA=180^{\circ }-\angle B=180^{\circ }-\angle D=\angle QDC$$ and, similarly, $\angle PAB=\angle QCD$. Since $AB=CD$, the triangles $PAB$ and $QCD$ are congruent with the same orientation. Moreover, $PQE$ is isosceles with $EP=EQ$.



In Solution 1 we have proved that triangles $IAB$ and $ICD$ are also congruent with the same orientation. Then we conclude that quadrilaterals $PBIA$ and $QDIC$ are congruent, which implies $IP=IQ$. Then $EI$ is the perpendicular bisector of $PQ$ and, therefore, $EI\perp PQ⟺EI\perp BC$.