Question

In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle in D. If ∠ACD = y^o and ∠AOD = x°, prove that x = 3y.

Answer 1

It is given that AB is a chord of a circle with centre O and AB is produced to C such that BC = OB

We know that ∠BOC and ∠BCO form isosceles triangle

∠BOC = ∠BCO = y^o

We know that OA and OB are the radii of same circle

OA = OB

We know that ∠OAB and ∠OBA form isosceles triangle

∠OAB = ∠OBA = 2y^o

From the figure we know that

Exterior ∠AOD = ∠OAC + ∠ACO

It can be written as

Exterior ∠AOD = ∠OAB + ∠BCO

So we get

Exterior ∠AOD = 3y^o

It is given that

∠AOD = x^o

So we get

x^o = 3y^o

Therefore, it is proved that x = 3y.