Question

Prove that the line joining the centres of two intersecting circles is the perpendicular bisector of the line joining the points of intersection.

Answer 1

AC = AD (Radii of the same circle)

AE = AE (Common side)

BC = BD (Radii of the same circle)

ΔABC = ΔABD (Three sides are equal)

In equal triangles, angles opposite to equal sides are equal.

So, ∠CAE = ∠DAE

Consider ΔCAE and ΔEAD.

∠CAE = ∠DAE

AC = AD (Radii of the same circle)

AE = AE (Common side)

ΔAEC = ΔAED (Two sides and the angle between them) 

In equal triangles, sides opposite to equal angles are equal. 

So, CE = DE (∠CAE = ∠DAE) 

CE = DE ………(1) 

In equal triangles, angles opposite to equal sides are equal. So, ∠AEC = ∠AED 

∠AEC + ∠AED = 180° (Linear pair) 

∠AEC = ∠AED = 90° ………(2) 

From equation (1) and (2) 

The line joining the centres of the circles is the perpendicular bisector of the chord.