Hello,
There was a question in geometry which was marked as a simple question as follows(Diagram below):-
The diagonal AC of a quadrilateral $CDAB$ bisects the $\angle DCB$. It is given that $BC=BD$. $\angle CAB = 80$, $\angle DBC = 20$, then the measure of $\angle ADC$ in degree is......
Lets take the point of intersection of the two diagonals as O. Now since $\triangle DCB$ is isosceles, $\angle D$ (Of the triangle) + $\angle C + 20 = 180$. Simplifying, we get $\angle CDA = 80$, and thus $\angle ACD$ and $\angle ACB = 40.$ Now solving for $ \triangle ACB$ (Taking the whole of $\angle C$ as $20+x$) we get $x = 40$. Now $\angle O$ in $\triangle AOC = 120$ and using vertically opposite angles formula, i got the value of $\angle AOB$ and $DOC = 60$. Now i'm stuck here. From my diagram i can see that $\angle CAB$ and $\angle BDC$ are equal to each other And $\angle ABD$ and $\angle DCA$ are equal to each other. I thought of assuming $\angle BDA = \angle ACB$ but i have no formal proof for that.
ANY HELP IS APPRECIATED
As $\angle CDB=\angle CAB$, quadrilateral $ABCD$ is cyclic. It follows that $\angle ADB=\angle ACB=40°$.