
from definition of a tangent, orthogonality of points A and B is trivial.
We are to show that the other two points of intersection A' and B' of the circle with diameter AB with the original two circles are such that OA' and OB' are tangents where O is the centre of the third circle.
Let P be the centre of one of the circles whose tangents are drawn OA and OA'. Here triangle OAP and OA'P are congruent by S-S-S test.
Hence angle OAP = angle OA'P=90 degrees.
Thus, OA' is also a tangent which completes the proof.