Two circles intersect at two points $A$ and $B$. $AD$ and $AC$ are diameters to the two circles (see Fig.). Prove that $B$ lies on the line segment $DC$.

(N/A) Join $AB.$
$\angle ABD = 90^\circ$ (Angle in a semicircle is a right angle).
$\angle ABC = 90^\circ$ (Angle in a semicircle is a right angle).
Adding these two equations,we get:
$\angle ABD + \angle ABC = 90^\circ + 90^\circ = 180^\circ.$
Since the sum of the angles is $180^\circ$,the points $D, B,$ and $C$ form a straight line. Therefore,$B$ lies on the line segment $DC.$