TITLE: Geometry problem, with circle.
QUESTION [5 upvotes]: We have that $DS \parallel BA$ and $∠DOS+∠DTS=180°$ and $O$ is the centre of the circle. I should somehow prove that $AB=AC$. In case you'd tell me I didn't show any effort I want to tell you that I tried to make this drawing for 40 minutes. Thank you for any help you could provide me.

REPLY [3 votes]: If $AB=AC$, using properties of an isosceles triangle,
$$\angle ACB=\angle ABC$$so this will be the goal of our proof.

Some headstart for you:
We have
$$∠DOS+∠DTS=180^\circ$$
so
$$∠TDO+∠TSO=180^\circ\tag{1}$$
Since $DS\parallel BA$,
$$∠SDB+\underbrace{∠DBA}_{=\angle ABC}=180^\circ\tag{2}$$
We also have
$$∠ABC+∠ACB+∠BAC=180^\circ$$
Your turn :)