I and O are incentre and circumcentre of a `Delta ABC`. AI is produced and it intersects the circumcircle at D. `angle ABC=x, angle BID=y & angle BOD=z`. Find `(z+x)/(y)`

Answer

To solve the problem, we need to find the value of \((z + x)/y\) given the angles in triangle \(ABC\) where \(I\) is the incenter and \(O\) is the circumcenter. Let's break down the steps:

### Step 1: Understand the Angles
We are given:
- \(\angle ABC = x\)
- \(\angle BID = y\)
- \(\angle BOD = z\)

### Step 2: Relate the Angles
Since \(I\) is the incenter, it divides the angle \(A\) into two equal parts. Therefore:
\[ \angle AIB = 90^\circ + \frac{x}{2} \]
This is because the angle at the incenter is equal to \(90^\circ\) plus half of the angle opposite to it.

### Step 3: Use the Exterior Angle Theorem
In triangle \(BID\), we can apply the exterior angle theorem:
\[ \angle BOD = \angle BID + \angle AIB \]
Substituting the known values:
\[ z = y + \left(90^\circ + \frac{x}{2}\right) \]

### Step 4: Rearranging the Equation
Rearranging the equation gives us:
\[ z = y + 90^\circ + \frac{x}{2} \]

### Step 5: Express \(z + x\)
Now, we can express \(z + x\):
\[ z + x = y + 90^\circ + \frac{x}{2} + x \]
This simplifies to:
\[ z + x = y + 90^\circ + \frac{3x}{2} \]

### Step 6: Substitute into the Expression
Now, we need to find \(\frac{z + x}{y}\):
\[ \frac{z + x}{y} = \frac{y + 90^\circ + \frac{3x}{2}}{y} \]
This can be simplified to:
\[ \frac{z + x}{y} = 1 + \frac{90^\circ}{y} + \frac{3x}{2y} \]

### Step 7: Conclusion
Thus, the expression \(\frac{z + x}{y}\) can be expressed in terms of \(x\) and \(y\).

### Final Answer
The final expression is:
\[ \frac{z + x}{y} = 1 + \frac{90^\circ}{y} + \frac{3x}{2y} \]