Finding angles in quadrilateral.

Question

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.

Answer 1

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: $$ AL<AB+BL=AB+CD=AC.\tag1 $$

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:

$$\angle B=110^\circ,\quad \angle C=70^\circ.$$