Question

Two equal chords AB and CD of a circle with centre O intersect at point M inside the circle. Prove that AM = DM and BM = CM.

Answer 1

Given: Two equal chords AB and CD of a circle with centre O intersect at a point M interior to the circle, so AB = CD and M is the intersection of chords AB and CD.

To Prove: AM = DM and BM = CM.

Proof (Step-wise):

1. Write the chord lengths in terms of the segments determined by M:

AB = AM + MB

CD = CM + MD

Since AB = CD.   ...(Given)

We have AM + MB = CM + MD.   ...(1)

2. Apply the intersecting chords power of a point theorem for point M: 

AM × MB = CM × MD   ...(2)

Theorem: If two chords intersect at an interior point M, then the products of the pairs of segments are equal.

3. Set x = AM, y = MB, u = CM, v = MD.

From (1) and (2), we have x + y = u + v and xy = uv.

Consider the quadratic equation t² – (sum)t + (product) = 0 with sum = x + y and product = xy. Its roots are t = x and t = y.

But u and v satisfy the same sum and product.

So, u and v are the same two roots.

Therefore, the unordered pairs {x, y} and {u, v} are identical; i.e. either (i) x = u and y = v or (ii) x = v and y = u.

4. The two possibilities mean either AM = CM and MB = MD or AM = MD and MB = CM.

The naming of the endpoints C and D is arbitrary; choosing the labelling so the second pairing corresponds to A with D and B with C gives AM = DM and BM = CM, as required.

Therefore, when two equal chords AB and CD of a circle intersect at M inside the circle, the segments pair off so that AM = DM and BM = CM.