[AHSME 1971] Quadrilateral $ABCD$ is inscribed in a circle with diameter $AD = 4$. If sides $AB$ and $BC$ each have length $1$, then find $CD$.
I tried looking through the solution of this, and couldn't understand the following part:
Since $AB = BC$, arc $AB$ = arc $BC$, so $\angle BDA = \angle BDC$
Why is $\angle BDA = \angle BDC$?
Let $O$ (origin) be the midpoint of $AD$ and add the lines $CO$ and $BO$. Hopefully you will agree that $\angle COB = \angle BOA$ and thus $\angle COA$ is twice that of $\angle BOA$. There's a property of triangles in the circle that $\angle BDA$ is half of $\angle BOA$ and, similarly, $\angle CDA$ is half of $\angle COA$. Rearrange to get your desired equality.
