Answer :


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 = 2ADB ------------ (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,


3ACB = 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.


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