Question

circles described on two sides of a triangle as diameters meet at the midpoint of the third side prove that the triangle is isosceles​

Answer 1

Two circles, unless they are concentric (not the case here) or touch each other (in which case the points A, C and B need to be collinear which is not the case), will intersect at two points. In this case, the circles D and E intersect at C (the point common to their diameters BC and AC respectively). So, there is only one other point of intersection. Let that point be F.

Join AF and BF.

Necessary condition:

Now, as F is on circle D, hence ∠BFC=90∘ ( the angle subtended by the diameter of a circle at the circumference is 90∘). Similarly, F is on circle E, hence ∠AFC=90∘.

As ∠AFC+∠BFC=90∘+90∘=180∘, the points A, F and B are collinear, and hence F lies on AB, the third side.

Sufficient Condition:

Let the circle D intersect AB at F. Then ∠BFC=90∘. Hence, ∠AFC=90∘. But AC is the diameter of the circle E. Hence, the point F lies on the circle E (Follows from the principle that if the diameter of a circle subtends an angle of 90∘ at a point, then the point necessarily lies on the circumference of that circle — FE=AC2). As F is common to both the circles, it is a point of intersection, and lies on the third side AB.