Let the line through $F$ perpendicular to $BD$ and the line through $G$ perpendicular to $AC$ intersect at $P$. We want to prove that $PE\perp AD$. Let $F'=FP\cap BD$, $G'=GP\cap AC$, and $PE\cap AD=K$. Claim: $F'G'\parallel BC$ Proof: It suffices to show that $\frac{BE}{CE}=\frac{BF'}{CG'}$. However, $\frac{BF'}{CG'}=\frac{BF\cos\angle FBF'}{CG\cos\angle GCG'}=\frac{BF}{CG}=\frac{AB}{CD}$. Also, $\frac{BE}{CE}=\frac{\sin\angle BCE}{\sin\angle CBE}$ from LoS on $\triangle BCE$, but from LoS on $\triangle BAC$ and $\triangle BDC$, we have $\sin\angle BCE=\frac{AB\sin\angle BAC}{BC}$ and $\sin\angle CBE=\frac{CD\sin\angle BDC}{BC}$, so $\frac{\sin\angle BCE}{\sin\angle CBE}=\frac{AB}{CD}$, as desired.$\square$ Let $H=GG'\cap BC$. Note that $\angle PF'E=\angle PG'E=90^{\circ}$, so $PF'EG'$ is cyclic. Finally, $\angle DEK+\angle EDK=\angle F'EP+\angle BDA=\angle F'G'P+\angle BCA=\angle G'HC+\angle HCG'=90^{\circ}$, so $\angle DKE=90^{\circ}\implies PE\perp AD$, as desired.

Is this sketch valid? The problem actually holds for any $F$ and $G$ such that points $F$ and $G$ are "similar" wrt triangles $EBA$ and $ECD$. The point is that if you vary the two that way, then the second and third lines move linearly, and then their intersection moves with degree $\leq 2$. Then, you can check the problem for $3$ points. Easy ones are $F$ and $G$ lying on the perpendicular bisectors of $BE$ and $CE$, respectively; the points such that point of concurrence is $E$; $F=A$ and $G=D$