Let $$ABCD$$ a quadrilateral, $$AC\cap BD=\{O\}$$, $$\angle A=110^{\circ}$$, $$\angle DOC=60^{\circ}$$ and $$AC=AB+CD=BD$$.

Find $$\angle B$$, $$\angle C$$.

I tried a lot of constructions. I think $$ABCD$$ is a trapezoid.

• You could improve your question by explaining what constructions you tried, and explaining why you think ABCD is a trapezoid. Commented Jun 20, 2020 at 13:06

You are correct. The quadrilateral $$ABCD$$ is an isosceles trapezoid.
Draw through the point $$C$$ a line $$\ell_1$$ parallel to $$(BD)$$. Draw through the point $$B$$ a line $$\ell_2$$ parallel to $$(CD)$$. Let $$\ell_1\cap\ell_2=L$$. The quadrilateral $$BDCL$$ is by construction parallelogram, hence $$BL=CD$$, $$CL=BD$$.
Assume now that $$CD\not\parallel AB$$ (as done in the figure). From the triangle inequality we have: $$AL
On the other hand it follows from $$AC=CL$$, $$\angle ACL=60^\circ$$ that the triangle $$ACL$$ is equilateral. This means: $$AL=AC.\tag2$$ Since (1) and (2) cannot hold simultaneously, the assumption $$CD\not\parallel AB$$ was false.
In the case $$CD\parallel AB$$ no contradiction arises and it can be checked that the internal angles of the resulting isosceles trapezoid can take any value between $$60^\circ$$ and $$120^\circ$$. Thus the quadrilateral in question is constructible and the answer is:
$$\angle B=110^\circ,\quad \angle C=70^\circ.$$