Question

Two chords AB and CD of a circle with centre O intersect at point M such that ∠OMD = ∠OMA. Prove that AB = CD.

Answer 1

Given:Two chords AB and CD of a circle with centre O intersect at point M and ∠OMA = ∠OMD.

To Prove: AB = CD.

Proof [Step-wise]:

1. Consider the reflection r across the line OM.

Reflection r fixes every point on OM; in particular, r(O) = O and r(M) = M and r is an isometry that preserves distances.

2. The reflection r sends the ray MA to the ray symmetric to MA about OM.

Because ∠OMA = ∠OMD (given), the symmetric of MA about OM is the ray MD.

Thus, r(A) lies on MD.

3. Also OA = radius = OD and r preserves distances from O.

So, r(A) lies on the circle. The only point on the circle that lies on ray MD and is at distance OA from O is D.

Hence, r(A) = D.

4. Since r is an isometry, it sends the whole chord AB to the chord r(A)r(B) = Dr(B).

Because r maps the intersection point M to itself and r(A) = D, the image chord r(A)r(B) must be CD. 

Thus, r(B) = C.

5. Isometries preserve lengths.

So, AB = r(AB)

= r(A)r(B)

= DC = CD

Therefore, AB = CD.