Question

AB and CD are two equal chords of a circle with centre O, intersecting each other outside the circle at point M.

Prove that:

1. AM = CМ
2. BM = DM

Answer 1

Given: AB and CD are two equal chords of a circle with centre O; the two chords extended meet at an external point M, as in the figure.

To Prove:

1. AM = CM. 
2. BM = DM.

Proof [Step-wise]:

1. Put notation:

Let AM = x, MB = y, CM = u and MD = v. 

All these are positive lengths; by the picture the order on each secant is ABM and CDM.

So, x > y and u > v.

2. Power of a point M for two secants through M gives MA × MB = MC × MD. 

i.e., x × y = u × v.   ...(Equation 1)

3. Express chord lengths in terms of these segments.

On the secant ABM, we have AM = AB + BM.

So, AB = AM – BM = x – y.

Similarly, CD = CM – MD = u – v.

Since AB = CD (given), x – y = u – v.   ...(Equation 2)

4. From (2) rearrange:

x – u = y – v

Call this common difference s.

So, x = u + s and y = v + s.   ...(s may be positive, zero, or negative a priori)

5. Substitute x = u + s and y = v + s into (1):

(u + s)(v + s) = uv

⇒ uv + s(u + v) + s² = uv 

⇒ s(u + v) + s² = 0

⇒ s(s + u + v) = 0

6. u and v are positive lengths.

So, s + u + v = x + v > 0. 

Therefore, the factor (s + u + v) cannot be zero, hence s = 0.

7. s = 0 gives x – u = 0 and y – v = 0.

i.e., x = u and y = v. 

Thus, AM = CM and BM = DM.