A circle has center on the side $AB$ of the cyclic quadrilateral $ABCD$. The other three sides are tangent to the circle. Prove that $AD+BC=AB$.
There are a lot of solution available on Artofproblemsolving
But could someone give me a solution by extending $AD$ and $BC$? This is what I do when I tried to solve this problem, unfortunately I didn't reach my goal there. Here is what I could find: I let $K$ be the intersection of $AD$ and $BC$. Since $ABCD$ is cyclic we could get that $\angle{KDC}=\angle{CBA}=x$ and $\angle{KCD}=\angle{DAB}=y$. If I were able to find that $2x+y=180°$ then I'd be able to prove that $\triangle{KBA}$ is an isosceles triangle and $KB=AB$ and lastly prove that $KC=AD$ and we're done. But I'm stuck at finding $2x+y=180°$. Please help
Solution 7.
Let $O$ be a center of the circle, which tangent to the sides $AD,$ $DC$ and $BC$ and
let $E\in AB$ such that $AD=AE.$
Thus, it's enough to prove that $EB=BC$.
We can assume that $E$ is placed between $O$ and $B$.
Also, let $\measuredangle ADO=\measuredangle ODC=\alpha$ and $\measuredangle DCO=\measuredangle OCB=\beta.$
Thus, $$\measuredangle DEA=\frac{1}{2}(180^{\circ}-\measuredangle DAB)=\frac{1}{2}(180^{\circ}-(180^{\circ}-2\beta))=\beta=\measuredangle DCO,$$ which says that $DOEC$ is cyclic, which says: $$\measuredangle CEB=\measuredangle ODC=\alpha,$$ which since $\measuredangle EBC=180^{\circ}-2\alpha,$ gives $$\measuredangle ECB=\alpha,$$ $$EB=BC$$ and we are done!