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 $AB$.

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

• Please read descriptions of tags before using them - you used two tags (algebraic-geometry and euclidean-algorithm) which specifically said in their description that they were not applicable to this sort of problem. – KReiser Feb 5 at 0:10 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.
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 $$\angle C$$.