Question

AB and CD are two equal chords of a circle with centre O. If P and Q are mid-points of AB and CD, respectively.

Prove that:

1. ∠BPQ = ∠DQP
2. ∠APQ = ∠CQP

Answer 1

Given: AB and CD are equal chords of a circle with centre O. P and Q are the mid‑points of AB and CD, respectively.

To Prove:

1. ∠BPQ = ∠DQP
2. ∠APQ = ∠CQP

Proof [Step-wise]:

1. Since P and Q are mid‑points of chords AB and CD and O is the centre, the lines OP and OQ pass from the centre to the mid‑points of the chords.

By the standard circle theorem, a line from the centre to the midpoint of a chord is perpendicular to the chord.

Hence, OP ⟂ AB and OQ ⟂ CD.

2. Equal chords are equidistant from the centre.

As AB = CD, the perpendicular distances from O to the chords are equal.

So, OP = OQ.

3. In ΔPOQ

OP = OQ

So, ΔPOQ is isosceles with base PQ. 

Therefore, its base angles are equal: ∠OPQ = ∠OQP.

4. Because OP ⟂ AB, PB is perpendicular to OP.

Thus, the acute angle between PB and PQ equals 90° – ∠OPQ, i.e. ∠BPQ = 90° – ∠OPQ.

Similarly, since OQ ⟂ CD, QD ⟂ OQ.

So, ∠DQP = 90° – ∠OQP.

5. From step 3,

∠OPQ = ∠OQP 

So, from step 4: 

∠BPQ = 90° – ∠OPQ

= 90° – ∠OQP

= ∠DQP

This proves (i).

6. Points A, P, B are collinear, so ∠APQ and ∠BPQ are supplementary sum to 180°. 

Likewise, ∠CQP and ∠DQP are supplementary.

Since ∠BPQ = ∠DQP from (i), subtracting each from 180° gives ∠APQ = ∠CQP.

This proves (ii).

Therefore, (i) ∠BPQ = ∠DQP and (ii) ∠APQ = ∠CQP, as required.