Problem 1672

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 $A'$ be the point on $AC$ with $CA' = DE$. Since $AC$ is parallel to $DE$, $A'CDE$ is a parallelogram. Thus $\angle AA'E = \angle ACD$. It follows that $\angle AA'E = \angle A'AE$, as $\angle ACD = \angle CAE$. Further, the fact that $AH = CH - DE = CH - CA' = HA'$ guarantees that $H$ is the midpoint of the base $AA'$ of an isosceles triangle $EAA'$. Therefore $EH$ is orthogonal to $AA'$. 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 $\triangle BKC \sim \triangle HKG$,
$$\frac{BC}{GH} = \frac{BK}{HK} \quad (1)$$
Since $DE$ is parallel to $KH$, $\triangle BDE \sim \triangle BKH$. Thus
$$\frac{BK}{HK} = \frac{BD}{DE} \quad (2)$$
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 $DE$, $\triangle BKH \sim \triangle BDE$, which implies that
$$\frac{BK}{KH} = \frac{BD}{DE} \quad (3)$$
Since $GH \cdot BD = BC \cdot DE$,
$$\frac{BD}{DE} = \frac{BC}{GH} \quad (4)$$
Combining (3) and (4), we obtain $\frac{BK}{KH} = \frac{BC}{GH}$.
Let $G'$ be the intersection point of $KD$ and the circumcircle of the triangle $BHC$. Then $\triangle BCK \sim \triangle HG'K$, and so
$$\frac{BK}{KH} = \frac{BC}{G'H}$$
But, since $\angle BKC > 90^\circ$, there is only one point $X$ satisfying $\frac{BK}{KH} = \frac{BC}{XH}$. Thus $G' = G$, 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 $D'$ 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
$$CH = CD' + D'H = CD' + DE = AH + DE.$$
This proves the necessity. $\square$