What are the steps to prove that if congruent chords are equidistant from the centers of circles, then the circles are congruent?

Answer 1

Best to draw this along with me.

Consider chrd AB in circle V and chord BC in circle W.

We are given that measure AB = CD

Draw from center V and center W a radius VX and a radius WY perpendicular to AB and CD respectively, VX meeting AB at P and VY meeting CD at Q..

Remember that a radius perpendicular to a chord is the perpendicular bisector of the chord

[ sub-proof if needed] VA and VB are radii so VA = VB thus the triangle VAB is isosceles.

Va = VB, VP = VP, and angle VPA = angle VPB = 90 degrees by construction. Triangle VPA is congruent to triangle VPB by hypotenuse-leg. PA = PB as corresponding parts of congruent triangles. So VP or VX is the right bisector. The same in the second triangle]

The distances from the center are the measures of VP and WQ. We are given that these are equal so measure VP = WQ

We now have VP = WQ, AP = CQ = 1/2 AB = 1/2 CD, and angle VPQA = angle WQC = 90 degrees, so triangle VPA is congruent to triangle WQC by SAS

Therefore radius of circle V = VA = WC = radius of circle W as corresponding parts of congruent triangles.

Two circles with equal radii are congruest, so circle V is congruent to circle W.