$[ADA']=[BDB']\Longleftrightarrow\frac{AD}{BD}=\frac{\sin\alpha\cos\alpha}{\sin\beta\cos\beta}$. It easy to get $\angle AB'D=\angle DA'B=x$. Now use sine law for $\triangle ADB'$ and $\triangle BDA'$ to get $\frac{AD}{BD}=\frac{\sin(\beta+x)\cos\alpha}{\sin(\alpha+x)\cos\beta}$. Now we get $\frac{\sin(\beta+x)}{\sin(\alpha+x)}=\frac{\sin\alpha}{\sin\beta}$ and hence $x=\gamma$ so $\angle A'DB'=\gamma$.

Assume circle $(K) \equiv \odot(A'B'D)$ touches $AB$ at $D$. Let $BC, AC$ meet $(K)$ again at $E, F,$ resp. Since $A'FB'E$ is cyclic and $A'B'$ is antiparallel of $AB$ WRT $\widehat C$ $\implies$ $EF \parallel AB$. Powers of $A, B$ to $(K)$ are in ratio ${\frac{\overline{AD}^2}{\overline{BD}^2}} = \frac{\overline{AB'} \cdot \overline{AF}}{\overline{BA'} \cdot \overline{BE}} = \frac{\overline{AB'} \cdot \overline{AC}}{\overline{BA'} \cdot \overline{BC}} = \frac{b \cos \widehat A}{a \cos \widehat B}$. In addition, $\measuredangle CB'D = \measuredangle CB'A' + \measuredangle A'B'D = \measuredangle ABC + \measuredangle A'DB = \measuredangle DA'C$. Assume areas of $\triangle ADA', \triangle BDB'$ are equal. Then $1 = \frac{\frac{1}{2} [AD] \cdot [AA'] \cdot \sin \widehat{A'AD}}{\frac{1}{2} [BD] \cdot [BB'] \cdot \sin \widehat{B'BD}} = \frac{[AD] \cdot ([AB] \cdot \sin \widehat B) \cdot \cos \widehat B}{[BD] \cdot ([AB] \cdot \sin \widehat A) \cdot \cos \widehat A} = \frac{[AD] \cdot b \cos \widehat B}{[BD] \cdot a \cos \widehat A}$ $\implies$ $-\frac{\overline{AD}}{\overline{BD}} = \frac{a \cos \widehat A}{b \cos \widehat B}$

Assume both conditions are satisfied. Then ${\frac{b \cos \widehat A}{a \cos \widehat B} = \frac{\overline{AD}^2}{\overline{BD}^2}} = \frac{a^2 \cos^2 \widehat A}{b^2 \cos^2 \widehat B}$ $\implies$ $\frac{a^2}{b^2} = \frac{b \cos \widehat B}{a \cos \widehat A} = -\frac{\overline{BD}}{\overline{AD}}$ $\implies$ $CD$ is also C-symmedian of $\triangle ABC$. Quadrilateral $CB'DA'$ has equal opposite angles at $A', B'$ and diagonal $CD$ (C-symmedian) cuts the other diagonal $A'B'$ (antiparallel of $AB$ WRT $\widehat C$) at its midpoint $\implies$ $CB'DA'$ is either parallelogram or kite. But if $CB'DA'$ was kite $\implies$ $ [A'C'] = [CB']$ $\implies$ $[BC] = [CA]$, which contradicts problem condition. $CB'DA'$ is therefore parallelogram $\implies$ $\measuredangle B'DA' = \measuredangle A'CB' \equiv \measuredangle BCA$.

since $ABA'B'$ is cyclic $\angle BA'B'=180-\angle BAC$ so now since $\angle ADB'=\angle DB'A'$ we get $\angle AB'D=\angle BA'D$ however thanks to the equal areas the heights of $DB'A$ and $A'BD$ from $B'$ and $A'$ with length $p,q$ satisfy $p/q=AD/BD$. however only 2 points on circle $AB'D$ are distant from $AB$ by exactly $p$ and those 2 points both with $A,B$ make similar triangles to $BA'D$ so $B'$ is one of those 2 points so $AB'D\sim A'BD$ and $AC\not = BC$ than $\angle ADB'=\angle ABC$ meaning $B'D||BC$ similarly $A'D||AC$ so $\angle A'DB=\angle ACB$

Outlines of my ugly proof Let $\angle {A'DB'}=\theta$ and $AC \cap \odot A'B'D=X$.Then $[AA'D]=[BB'D] \Rightarrow BD*BB'cosA=AD*AA'cosB \Rightarrow \frac{AD}{BD}=\frac{sin2C}{sin2B}$.Substitute $BD=c-AD$ to get $AD$.By power of point theorem we have $AX=\frac{AD^2}{AB'}$.Hence get length $AX$ and hence $B'X$.But sine rule in $\triangle B'A'X$ gives $B'X=\frac{B'A'sin(B-\theta)}{sin \theta}$.Substitute $B'A'=Rsin2C$ and thus equate the two equations for $B'X$ and get $\theta=\angle C$.

Good problem, although easy bashable.

Wolowizard, your solution is painfully wrong. Your last implication is incorrect and in fact you just made the first step towards the solution. Here is mine solution: From the areas equality we easily get $\frac{AD}{BD}=\frac{\sin \alpha \cos \alpha}{\sin \beta \cos \beta}$ (we can do it Wolowizard's way or intersect $A'B'$ and $AB$). Now we denote $\angle ADB'=x$ and $\angle BDA'=y$. Now we notice that $\alpha + x=\beta +y$. Now by applying sine theorems in $\triangle ADB'$ and $\triangle BDA'$ we get $\frac{\sin \alpha \cos \alpha}{\sin \beta \cos \beta} =\frac{AD}{BD}=\frac{\cos \alpha \sin y}{\cos \beta \sin x}$. Now we conclude $\frac{\sin y}{\sin (\pi-x)}=\frac{\sin \alpha}{\sin (\pi - \beta)}$ so now because $\pi +y -x=\pi + \alpha - \beta$ we conclude $y=\alpha$ and $x=\beta$ so $\angle A'DB'=\angle ACB$ so we are finished.

My solution Let $\angle DA'A=x$,$\angle DB'B=y$. From Sine theorems on triangles $\triangle AA'D$ and $\triangle BB'D$ and from the surface equality condition we have $\frac{AD}{BD}=\frac{\sin{2a}}{\sin{2b}}$. Now since $AD$ and $BD$ touch circle $A'B'D$ we have that $\angle B'DA=\angle DA'B'$ which implies that $x=y$. Now from Sine theorem on triangle $\triangle BA'D$ we have that $\frac{BA'}{BD}=\frac{\sin{90-x+b}}{\sin{90-x}}$ which is equivalent to $\frac{\sin{a+b}\cos{a-b}}{\sin{b}}=\frac{\sin{90-x+b}}{\sin{90-x}}$ which implies that $\cot{90-x}=\cot{c}$ implying $x=90-c$ which implies $\angle A'DB'=c$.