Question

In the adjoining figure, AD is the diameter of the circle whose centre is O. If AB || CD, prove that AB = CD.

Answer 1

Given: AD is the diameter of the circle with centre O. AB || CD.

To Prove: AB = CD.

Proof [Step-wise]:

1. Let M and N be the mid-points of chords AB and CD, respectively.

2. The line from the centre to the midpoint of a chord is perpendicular to the chord.

Hence, OM ⟂ AB and ON ⟂ CD.

So, AM = MB and CN = ND.

3. Consider right triangles ΔOAM and ΔODN.

OA = OD   ...(Radii)

∠OMA = ∠OND = 90°   ...(From step 2)

Since AM lies on AB and DN lies on CD and AB || CD, the acute angles made by OA with AM and by OD with DN are equal; thus ∠OAM = ∠ODN.

4. Therefore, ΔOAM ≅ ΔODN by RHS congruence right angle, hypotenuse OA = OD and an acute angle equal.

5. From the congruence,

AM = DN

Multiply both sides by 2:

AB = 2·AM and CD = 2·DN

So, AB = CD.