56th International Mathematical Olympiad Shortlisted Problems

Since $HG\parallel AB$ and $BG\parallel AH$, we have $BG\perp BC$ and $CH\perp GH$. Therefore, the quadrilateral $BGCH$ is cyclic. Since $H$ is the orthocenter of the triangle $ABC$, we have $\angle HAC=90^{\circ }-\angle ACB=\angle CBH$. Using that $BGCH$ and $CGJI$ are cyclic quadrilaterals, we get $$\angle CJI=\angle CGH=\angle CBH=\angle HAC.$$ Let $M$ be the intersection of $AC$ and $GH$, and let $D\ne A$ be the point on the line $AC$ such that $AH=HD$. Then $\angle MJI=\angle HAC=\angle MDH$. Since $\angle MJI=\angle MDH$, $\angle IMJ=\angle HMD$, and $IM=MH$, the triangles $IMJ$ and $HMD$ are congruent, and thus $IJ=HD=AH$.

---

Alternative solution.

Obtain $\angle CGH=\angle HAC$ as in the previous solution. In the parallelogram $ABGH$ we have $\angle BAH=\angle HGB$. It follows that $$\angle HMC=\angle BAC=\angle BAH+\angle HAC=\angle HGB+\angle CGH=\angle CGB.$$ So the right triangles $CMH$ and $CGB$ are similar. Also, in the circumcircle of triangle $GCI$ we have similar triangles $MIJ$ and $MCG$. Therefore, $$\frac{IJ}{CG}=\frac{MI}{MC}=\frac{MH}{MC}=\frac{GB}{GC}=\frac{AH}{CG}.$$ Hence $IJ=AH$.