Each of two equal circles passes through the centre of the other and the two circles intersect each other at the points A and B. If a straight line through the point A intersects the two circles at points C and D, let us prove that ΔABCD is an equilateral triangle.

As the circles with centre X and Y are equal, the radius of both the circles are equal.

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.

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

Since AXBC is a quadrilateral triangle, therefore the sum of opposite sides of a quadrilateral is equal to 180°

⇒ ∠AYB + ∠ACB = 180 --------- (2)

Also, by using above theorem, we get

⇒ ∠AYB = 2∠ADB ------------ (3)

In ΔAXB and ΔAYB,

AX = AY (radius of equal circles)

BX = BY (radius of equal circles)

AB = AB (common)

⇒ ΔAXB ≅ ΔAYB, by SSS congruency

∴ ∠AXB = ∠AYB ------------ (4)

From equation (1), (2) and (4) we get,

⇒ 3∠ACB = 180

⇒ ∠ACB = 60°

From equation (1), (3) and (4) we get,

∠ACB = ∠ADB = 60°

In ΔBCD, sum of all angles of a triangle is equal to 180°

⇒ ∠ACB + ∠ADB + ∠CBD = 180°

⇒ ∠CBD = 60°

As each angle in ΔBCD is equal to 60°, therefore ΔBCD is an equilateral triangle.