Let $ABCD$ be a cyclic convex quadrilateral such that $AD + BC = AB$. Prove that
the bisectors of the angles ADC and BCD meet on the line $AB$.
I tried to find similar triangles since the angles are getting bisected, however I couldn't go anywhere in that direction. I also tried seeing if there were any properties that could be useful about the cyclic quadrilateral. I found properties from here: https://www.quora.com/What-are-the-properties-of-a-cyclic-quadrilateral-with-images
On
Let $\angle D=2\delta$ and $\angle C=2\gamma$, and wlog assume $\delta\geqslant \gamma$ (to be consistent with your picture). Then $\angle A=180^\circ-2\gamma$ and $\angle B=180^\circ-2\delta$. Denote by $M$ the point on $AB$ such that $AM=AD$, so $BM=BC$ as well. Then $\angle ADM=\angle AMD=\gamma$ and $\angle BMC=\angle BCM=\delta$.
Denote by $N$ the intersection of $AB$ and the bisector of $\angle D$; since $\delta\geqslant\gamma$ we have $\mathcal B(A,M,N,B)$, and $\angle MDN=\delta-\gamma$. Since $\angle NMC=\angle BMC=\delta$ and $\angle NDC=\delta$ we get that $NCDM$ is cyclic, so $\angle MCN=\angle MDN=\delta-\gamma$. Now, $\angle BCN= \angle BCM-\angle NCM= \delta-(\delta-\gamma)= \gamma$, which means that $N$ is on the bisector of $\angle C$.
Let angle bisector for $\angle BCD$ meet $AB$ at $F$ (so we have to prove that $DF$ is angle bisector for $\angle ADC$), then $$\angle BCF = FCD = \alpha\;\;\;\;{\rm and }\;\;\;\;\;\angle BAD = 180^{\circ} -2\alpha$$ and let $E$ on $AB$ be such that $BE = BC$, then $AE = AD$ and $$\angle FED = \angle AED = \angle ADE = \alpha$$ so $\angle FED = \angle FCD = \alpha $ and thus $CDFE$ is cyclic!
Now, if $\angle FDC = \beta$ then $\angle BEC = \angle ECB = \beta$, so $$\angle 180^{\circ} -2\beta \implies \angle ADC = 2\beta$$ and thus $DF$ is also angle bisector but for $\angle ADC$ and we are done.