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
Best Answer
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.