In the original statement it was said about cyclic quadrilateral. Without it the problem is incorrect. By additional assumption we have $\angle ABD=\angle ACD=\angle ADC$. Let $F$ be the reflection of $D$ over $E$. Obviously $$DE-BE=FE-BE=BF.$$We have $$\angle AFB=\angle ADB=\angle ACB,\ \angle ABF=180^\circ-\angle ADC=\angle ABC.$$Therefore $$\triangle ABF\sim \triangle ABC,$$but they share a common side $AB$ so $$\triangle ABF\equiv \triangle ABC$$and $$BF=BC.$$QED

We have $\angle ADC = \angle DBA = \angle ACD$, so $\triangle ACD$ is isosceles and thus $A$ is the midpoint of arc $\widehat{CBD}$. Then by Archimedes' Theorem, $DE=BE+BC$, so $BC=DE-BE$.

Motivated by the fact that a difference of lengths is annoying to deal with... First, easy angles give us that $AC=AD$. Apply the Law of Cosines on $\triangle ABC$. \[AC^2=AB^2+BC^2-2AB\cdot BC\cos\angle ABC.\]Using the fact that $ACD$ is isosceles, we turn this into \[AC^2=AB^2+BC^2+2AB\cdot BC\cos\angle ADC=AB^2+BC^2+AB\cdot BC\cdot\frac{CD}{AD}.\]By Ptolemy's Theorem, \[AB\cdot CD=AC\cdot BD-BC\cdot AD.\]Plugging this in and using $AC=AD$ gives \[AC^2=AB^2+BC^2+BC(BD-BC)=AB^2+BC\cdot BD.\]Hence, $BD(DE-BE)=DE^2-BE^2=AD^2-AB^2=AC^2-AB^2=BC\cdot BD$. So, $BC=DE-BE$, as desired.

Let angle ADB=angle DBA=x.ABCD is cyclic, Let angle BCD=y So, angle BDC=2x-y angle ADB=y-x angle ABD=x In ∆ ADB, sin x/ cos x=AE/BE In ∆ADE, sin(y-x) / cos(y-x)= AE/DE DE/BE=2sin x cos(y-x) / 2 sin(y-x) cos x = sin y+ sin(2x-y) / sin y- sin(2x-y) DE+BE / DE-BE=sin y/ sin (2x-y) BD / DE-BE=sin y/ sin (2x-y) DE-BE/ BD =sin(2x-y) / sin y In ∆ BDC, by sine rule , BC/ BD= sin(2x-y)/ sin y BC/ BD=DE-BE / BD So, DE-BE=BC ( Proved) # Krishijivi

ADC=DBA=DCA -> ACD is isosceles with AC=AD Draw altitude AF inside ACD, clearly ABE~ACF -> AB*CF=BE*AC By Ptolemy's, AB*CD+AD*BC=AC*BD AC(BD-BC)=AB*CD=2*AB*CF=2*BE*AC BD-BC=2*BE BE+ED=2*BE+BC BC=DE-BE.