Question

extension
here are two circles of different sizes that intersect in 2 places,
with centers at a and b and points of intersection at c and d.
prove that segment ab must be perpendicular to segment cd

Explanation:

Step1: Connect AC, AD, BC, BD

Since \(AC = AD\) (radii of circle \(A\)), \(\triangle ACD\) is isosceles. Similarly, \(BC=BD\) (radii of circle \(B\)), \(\triangle BCD\) is isosceles.

Step2: Use the property of isosceles triangles

Let \(M\) be the intersection point of \(AB\) and \(CD\). In \(\triangle ACD\), the line \(AB\) is the axis of symmetry of \(\triangle ACD\) (because \(A\) is equidistant from \(C\) and \(D\)). In \(\triangle BCD\), \(AB\) is the axis of symmetry of \(\triangle BCD\) (because \(B\) is equidistant from \(C\) and \(D\)).
In an isosceles triangle, the axis of symmetry is also the altitude. So, \(AB\perp CD\)

Answer:

Segment \(AB\) is the perpendicular bisector of segment \(CD\) because \(A\) is equidistant from \(C\) and \(D\) ( \(AC = AD\), radii of circle \(A\)) and \(B\) is equidistant from \(C\) and \(D\) (\(BC = BD\), radii of circle \(B\)). By the property of isosceles triangles (the line joining the vertex (or in this case, the centers \(A\) and \(B\)) to the mid - point of the base (the mid - point of \(CD\)) is perpendicular to the base), \(AB\perp CD\)