Question

O is the centre of the circle.If BC=OB,prove that x=3y.

Answer 1

Given: BC=OB 

To prove: x = 3y

Proof:
∠BOC = ∠BCO = y (Since. BC=BO)

Now, In Triangle OBC,
∠ ABO = ∠BOC + ∠BCO (Exterior angle of a triangle is equal to the interior opposite angles) 
Therefore, 
∠ABO = 2y

Again, 
∠OBA = ∠ OAB = 2y
[Since AO = BO (since they are radii of same circle)]

In Triangle AOB,
∠ABO + ∠BAO + ∠BOA = 180 ( Angle sum property of triangle)
2y + 2y + ∠BOA = 180
∠BOA = 180 - 4y 

Now, 
∠AOC + x = 180 (linear pair)
180-4y + y + x = 180 
180 - 3y + x = 180
x = 180-180 + 3y 

Therefore, x = 3y

Answer 2

Answer:

Step-by-step explanation:

Given OB=BC, then from ΔOBC we get, ∠BOC=∠BCO=y. [ They are corresponding to the equal sides]

Then ∠OBC=180

o

−2y and OBA=180

o

−(180

o

−2y)=2y.

In ΔABO we have, OB=OA [ Equal to radius of the circle]

Then ∠OAB=∠OBA=2y. [ They are corresponding to the equal sides]

Then ∠AOB=180

o

−4y.

Now,

∠AOC=180

o

−4y+y=180

o

−3y.

Then,

x=180

o

−(180

o

−3y)=3y