I drew a pretty diagram for you.
Let
![](https://latex.codecogs.com/png.latex?\bg_white \angle ACD=x, \angle CDB=y)
, also let the intersection of the chords be
![](https://latex.codecogs.com/png.latex?\bg_white \angle ACD= \angle ABD= x)
(Angles at the circumference subtended by the same arc.)
Similarly,
Hence,
![](https://latex.codecogs.com/png.latex?\bg_white AB=CD)
(Corresponding sides of congruent triangles.)
Let
![](https://latex.codecogs.com/png.latex?\bg_white \ANGLE CBE = a)
(Angle at the circumference subtended by the same arc.)
Similarly,
Now,
![](https://latex.codecogs.com/png.latex?\bg_white AD=BC)
(Corresponding sides in congruent triangles.)
Now, quadrilateral
![](https://latex.codecogs.com/png.latex?\bg_white ABCD)
,
![](https://latex.codecogs.com/png.latex?\bg_white a+y+b+x=180)
(Opposite interior angles in a cyclic quadrilateral are supplementary.)
Now in
![](https://latex.codecogs.com/png.latex?\bg_white \Delta BAD)
and
![](https://latex.codecogs.com/png.latex?\bg_white \Delta BCD)
,
![](https://latex.codecogs.com/png.latex?\bg_white AD=BC)
(As proven.)
![](https://latex.codecogs.com/png.latex?\bg_white AB=CD)
(As proven.)
![](https://latex.codecogs.com/png.latex?\bg_white BD)
is common.
Hence,
![](https://latex.codecogs.com/png.latex?\bg_white \Delta BAD \equiv \Delta BCD)
,
Now,
![](https://latex.codecogs.com/png.latex?\bg_white a+y=b+x)
(Corresponding angles in congruent triangles.)
Back to our equation of,
Hence
![](https://latex.codecogs.com/png.latex?\bg_white BD)
is the diameter, angle at the semi circle is 90.