Proof:

Let's consider a triangle ABC with sides AB, BC, and AC. We will draw two circles taking sides AB and AC as diameters.

Step 1: Drawing the circles
1. Take side AB as the diameter and draw a circle with center O1.
2. Take side AC as the diameter and draw a circle with center O2.

Step 2: Establishing the intersection point
1. Let the circles intersect at point P.
2. To prove that P lies on side BC, we need to show that ∠BPC = 90°.

Step 3: Establishing the relationships
1. Since O1 is the center of the circle passing through A and B, we have ∠O1AB = 90°.
2. Similarly, since O2 is the center of the circle passing through A and C, we have ∠O2AC = 90°.
3. From the above two statements, we can conclude that ∠O1AB = ∠O2AC.

Step 4: Triangles and angles
1. Consider the triangle BPC.
2. We know that ∠O1AB = ∠O2AC (from step 3).
3. Also, ∠O1BA = ∠O2CA (as both are radii of their respective circles).
4. From the above two statements, we can conclude that ∠ABO1 = ∠ACO2.

Step 5: Completing the proof
1. Since ∠ABO1 = ∠ACO2, we have ∠ABP = ∠ACP.
2. Adding ∠ABP and ∠ACP, we get ∠BPC.
3. Therefore, ∠BPC = ∠ABP + ∠ACP = 180° - ∠BAP - ∠CAP (by triangle angle sum property).
4. Since ∠BAP + ∠CAP = 180° (as they are angles of a triangle), we have ∠BPC = 180° - 180° = 0°.
5. Hence, ∠BPC = 0°, which implies that P lies on side BC.

Therefore, the point of intersection of the circles lies on the third side of the triangle.