Let $F$ be the point on ray $DA$ such that $DF=DC$. Let $G$ be the intersection between $BD$ and $CF$. Claim 1: $BEFG$ is cyclic. Proof: By angle chasing, $$\angle GFD=\frac{180^o-\angle FDC}{2}=\frac{\angle ABC}{2}=\angle EBG \square$$ Claim 2: The intersection $H$ of $BE$ and $CF$ lies on the circle $(ABCD)$. Proof: Redefine $H$ as the intersection of $BE$ and $(ABCD)$. Then by angle chasing, $$\angle DCF=\angle DFC=\angle DBH=\angle DCH$$which shows that $H, F, C$ are collinear. $\square$ We now rewrite the required equality as follows: $$DE\times (AB+BC)=AC\times BD$$$$DE\times AB + DE\times BC = AB\times CD+AD\times BC$$$$BC\times (AD-DE)=AB\times (DE-DC)$$$$BC\times AE=AB\times (DE-DF)$$$$BC\times AE=AB\times EF$$ This is true because $$\frac{EF}{BC}=\frac{EF}{BG}\times \frac{BG}{BC}=\frac{HE}{HG}\times \frac{HG}{HD}=\frac{HE}{HD}=\frac{AE}{AB} \blacksquare$$

Let $BE\cap (ABCD)=F$, the angle bisector of $CBA$ meet $AC$ at $G$, and $CF$ meet $AD$ at $T$. By angle chasing, $\angle DCT=\angle DBF=\frac12\angle CBA=90-\frac12 \angle CDT$, hence $CD=TD$. Also, $\angle GAE=\angle CAD=\angle CBD=\angle GBE$, so $BAGE$ is cyclic. Thus, $\angle AGE=\angle ABE=\angle ACT$ so $GE\parallel CT$. By ratio, $\frac{AB}{BC}=\frac{AG}{GC}=\frac{AE}{ET}$ so $AB\times ET=AE\times BC$. $\implies AB\times(DE-TD)=(AD-ED)\times BC \implies (AB+BC)DE=BC\times AD+CD\times AB=AC\times BD$.