This problem is close to the problem in here

Let $(CDE)$ intersect $BC$ at $X$. Note that \begin{align*} BC^2&=BE\cdot BD+CD\cdot CA \implies BC^2-BX\cdot BC&=CD\cdot CA \\ CX\cdot CB=CD\cdot CA \end{align*}so $ABXD$ is cyclic. Then let $(BCE)$ intersect $(ADE)$ at $P\neq E$. We claim that $P$ is independent of $D$. Firstly, note that $\angle BPC=\angle BEC=180^\circ - \angle DXC=\angle BAD$ so $P$ lies on a fixed circle through $B$ and $C$. Furthermore, \[\angle PCB=\angle PEB=\angle PAC\]It follows that $P$ is the intersection point of a fixed circle through $B$ and $C$ with the circle through $A$ tangent to $BC$ at $C$. Since $P$ is independent of $D$ and always lies on $(ADE)$, it follows that $P$ is the required `fixed point'.

Wait. Isn't it exactly same with BMO 2017 SL G1? Because let $F=(AED)\cap AB$, then $BC^2=CD\cdot CA+BE\cdot BD=CD\cdot CA+BF\cdot BA$. So as $D$ varies on $AC$, $F$ varies on $AB$ and we want to show $(AFD)$ passes through fixed point other than $A$, which is exact thing that G1 asks. It is very absurd to see same problems as G1 and G2 on the same Shortlist.