Like the adjoining figure, we draw two circles with centres C and D which intersect each other at the points A and B. We draw a straight line through the point A which intersects the circle with centre C at the point P and the circle with centre D at the point Q.

Let us prove that (i) ∠PBQ = ∠CAD (ii) ∠BPC = ∠BQD

(i) In ΔACB and ΔADB,

By the theorem:-

The angle formed at the centre of a circle by an arc, is double of the angle formed by the same arc at any point on circle.

and

--------- (1)

In ΔPCA, ∠PAC + ∠CPA = 180 - ∠PCA

⇒ 2∠CAP = 180 - ∠PCA [As ∠PAC = ∠CPA]

Similarly, in ΔADQ,

⇒ 2∠QAD = 180 - ∠ADQ [As ∠QAD = ∠AQD]

Since, PAQ is a straight line,

⇒ ∠CAD = 180 – (∠CAP + ∠QAD)

On substituting values in above equation, we get,

------------ (2)

From (1) and (2), we get, ∠CAD = ∠PBQ ---------- (3)

Hence, proved.

(ii) Also, as ∠CBA = ∠CAB and ∠ABD = ∠BAD

⇒ ∠CBA + ∠ABD = ∠CAB + ∠BAD

⇒ ∠CAD = ∠CBD ------------- (4)

From equations (3) and (4), we get,

∠PBQ = ∠CBD

⇒ ∠PBQ - ∠CBD = 0

⇒ ∠PBC - ∠QBD = 0

⇒ ∠PBC = ∠QBD ---------------- (5)

As PC = CB and we know that angles opposite to equal sides are equal

⇒ ∠BPC = ∠PBC ----------------- (6)

As BD = DQ and we know that angles opposite to equal sides are equal.

⇒ ∠BQD = ∠QBD ----------------- (7)

From equations (5), (6) and (7)

∠BPC = ∠BQD