Korean Mathematical Olympiad Final Round

Since $AI=CI$, $\angle ACI=\angle CAI$, and so $\angle ACE=\angle CAD$. Since $ACDE$ is cyclic, $\angle CAD=\angle CED$. Hence $\angle ACE=\angle CED$, and so $AC$ and $DE$ are parallel to each other.

Now we will show that if $CH=AH+DE$, then $GH\cdot BD=BC\cdot DE$. Let $A^{\prime }$ be the point on $AC$ with $C{A}^{\prime }=DE$. Since $AC$ is parallel to $DE$, $A^{\prime }CDE$ is a parallelogram. Thus $\angle A{A}^{\prime }E=\angle ACD$. It follows that $\angle A{A}^{\prime }E=\angle A^{\prime }AE$, as $\angle ACD=\angle CAE$. Further, the fact that $AH=CH-DE=CH-C{A}^{\prime }=H{A}^{\prime }$ guarantees that $H$ is the midpoint of the base $A{A}^{\prime }$ of an isosceles triangle $EA{A}^{\prime }$. Therefore $EH$ is orthogonal to $A{A}^{\prime }$. It follows that $\angle BHC=90^{\circ }=\angle BGC$, and so the points $B$, $C$, $G$, $H$ are cyclic. Let $K$ be the intersection point of $BG$ and $CH$. Since $△BKC\sim △HKG$, $$\frac{BC}{GH}=\frac{BK}{HK}(1)$$ Since $DE$ is parallel to $KH$, $△BDE\sim △BKH$. Thus $$\frac{BK}{HK}=\frac{BD}{DE}(2)$$ Combining (1) and (2), we obtain $\frac{BC}{GH}=\frac{BD}{DE}$, and hence $GH\cdot BD=BC\cdot DE$. This proves the sufficiency.

Conversely, assume $GH\cdot BD=BC\cdot DE$. We will show that $CH=AH+DE$. Since $AC$ is parallel to $DE$, $△BKH\sim △BDE$, which implies that $$\frac{BK}{KH}=\frac{BD}{DE}(3)$$ Since $GH\cdot BD=BC\cdot DE$, $$\frac{BD}{DE}=\frac{BC}{GH}(4)$$ Combining (3) and (4), we obtain $\frac{BK}{KH}=\frac{BC}{GH}$. Let $G^{\prime }$ be the intersection point of $KD$ and the circumcircle of the triangle $BHC$. Then $△BCK\sim △H{G}^{\prime }K$, and so $$\frac{BK}{KH}=\frac{BC}{G^{\prime }H}$$ But, since $\angle BKC>90^{\circ }$, there is only one point $X$ satisfying $\frac{BK}{KH}=\frac{BC}{XH}$. Thus $G^{\prime }=G$, which implies that $B$, $C$, $G$, $H$ are cyclic. So, $\angle BHC=\angle BGC=90^{\circ }$. Since $AC$ is parallel to $DE$, $\angle DEH=90^{\circ }$. Let $D^{\prime }$ be the foot of the altitude from $D$ to the line $AC$. Since $H$ is the foot of the altitude from $E$ to the line $AC$, we have $$CH=C{D}^{\prime }+D^{\prime }H=C{D}^{\prime }+DE=AH+DE.$$ This proves the necessity. $\square$