I and 0 are respectively the in-centre and circumcentre of a triangle ABC. The line AI produced intersects the circumcircle of ΔABC at the point D. If ∠ABC = x°, ∠BID = y° and ∠BOD = z°, then $$\frac{z+x}{y}=$$

For cicumcentre and its chord BD

$$\angle$$BAD = $$\angle$$BOD/2

=> $$\angle$$BAD = z/2

In $$\triangle$$ABE

=> $$\angle$$BEA + $$\angle$$EAB + $$\angle$$ABE = 180

=> $$\angle$$BEA = 180 - z/2 - x

Since, BI is angle bisector

=> $$\angle$$IBE = $$\angle$$ABE/2

=> $$\angle$$IBE = X/2

Now, in $$\triangle$$IBE

=> $$\angle$$IBE + $$\angle$$BIE + BEI = 180

=> $$\frac{x}{2} + (180-\frac{z}{2} - x) + y$$ = 180

=> $$y = \frac{x}{2} + \frac{z}{2}$$

=> $$\frac{x+z}{y} = 2$$