this is my solution first for a triangle ABC we have B=2C <=> b ² =c ² +ca. (it can be proved easily) return to our problem AB+AD=CD and AC+AD=BC <=> b ² + bc + 2bccos(A/2) = ab+ac (1) and bc + c ² +2bccos(A/2) = ab (2) (1) - (2) and we have b ² = c ² + ca <=> B = 2C we can write (1) in another way (1) <=> 2bccos(A/2) = (a - b)(b + c) <=> sinAsinBsinC = (sinA - sinB)(sinA + sinB)sin(A/2) <=> sinAsinBsinC = 4cos((A+B)/2)sin((A-B)/2)sin((B+C)/2)cos((B-C)/2)sin(A/2) = 4sin(C/2)sin((A-B)/2)cos(A/2)cos(C/2)sin(A/2) (because B = 2C) <=> sin((A-B)/2) = sinB <=> A = 3B we also have B = 2C. So A = 120o B = 40o C = 20o

Are there any other solutions?(Not using trigonometry)?

Let $B'$ be the reflection of $B$ in $AD$, then easily we have $B'C=AC-AB=BC-CD=BD=B'D$, so $\angle B=\angle AB'D=2\angle C$. Meanwhile, let $E$ be the intersection of the perpendicular bisector of $AC$ and $BC$. Then $\angle AEB=2\angle C=\angle B$, thus $AB=AE=CE$, and hence $DE=CD-CE=CD-AB=AD$, so we have $\angle DAE=\angle DEA=2\angle C$, hence $\angle CAD=3\angle C$, and consequently we know that $\angle A=6\angle C$. Finally we have $\angle C=20^\circ$, $\angle B=40^\circ$,$\angle C=120^\circ$.