If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.

To prove that the centers of two circles that intersect at two points lie on the perpendicular bisector of the common chord, you can follow these steps:

Let's call the centers of the two circles O1 and O2, and the points where they intersect as A and B. Also, let's denote the common chord as AB.

Draw a line segment AB connecting the two points of intersection (the common chord).
Find the midpoint M of the line segment AB. The midpoint M is the point that divides AB into two equal parts.
Draw lines from O1 and O2 to the midpoint M, creating two line segments OM1 and OM2, where OM1 connects O1 to M, and OM2 connects O2 to M.
Now, we will prove that OM1 is perpendicular to AB and OM2 is perpendicular to AB:

Consider triangle O1AM. Since O1 is the center of the first circle, OA1 is the radius of the first circle, and AM is half of AB (since M is the midpoint of AB), we can write:

OA1 = AM (1)

Similarly, consider triangle O2BM. Since O2 is the center of the second circle, OB2 is the radius of the second circle, and BM is also half of AB, we can write:

OB2 = BM (2)

Now, we will show that triangle O1AM is congruent to triangle O2BM. Since the circles intersect at two points, OA1 and OB2 have the same length (they are both radii of their respective circles). So, from (1) and (2), we have:

OA1 = OB2

Additionally, AM and BM are both half of AB, which is a common line segment. Therefore, from side-side-side congruence criteria, we can conclude that:

Triangle O1AM is congruent to triangle O2BM.

Now, if two triangles are congruent, their corresponding angles are equal. In particular, angle O1MA is equal to angle O2MB.

Since O1MA and O2MB are corresponding angles in congruent triangles, they have the same measure. If two lines intersect, and their corresponding angles are equal, then these lines are perpendicular to each other.

Therefore, OM1 is perpendicular to AB, and OM2 is also perpendicular to AB. This proves that the centers of the two circles, O1 and O2, lie on the perpendicular bisector of the common chord AB.