Let ABC be a triangle in which the line joining the circumecentre and incentre is parallel to base BC of the triangle. Then answer the following questions :
If ODEI is a square where O and I stands for circumcentre and incentre, respectively and D and E are the point of perpendicular from O and I on the base BC, then

Options

A. `(r )/(R )=(3)/(8)`
B. `(r )/(R )=2-sqrt(3)`
C. `(r )/(R )=sqrt(2)-1`
D. `(r )/(R )=(1)/(4)`

Explanation

To solve the problem step by step, we will analyze the given conditions and derive the necessary equations.

### Step 1: Understand the Geometry
We have a triangle \( ABC \) with circumcenter \( O \) and incenter \( I \). The line \( OI \) is parallel to the base \( BC \). Points \( D \) and \( E \) are the feet of the perpendiculars dropped from \( O \) and \( I \) to line \( BC \).

**Hint:** Visualize the triangle and the positions of the circumcenter, incenter, and the points where the perpendiculars meet the base.

### Step 2: Recognize the Square
Since \( ODEI \) is a square, we have:
- \( OD = OI \)
- \( DE = OE \)

**Hint:** Recall that in a square, all sides are equal, and the diagonals are equal as well.

### Step 3: Use the Relationship Between \( OI \) and \( R, r \)
The distance \( OI \) can be expressed using the formula:
\[ OI = \sqrt{R^2 - 2Rr} \]
where \( R \) is the circumradius and \( r \) is the inradius.

**Hint:** This formula relates the circumradius and inradius of a triangle.

### Step 4: Express \( OD \)
The length \( OD \) can also be expressed as:
\[ OD = r \cos A \]
where \( A \) is the angle at vertex \( A \).

**Hint:** Remember that \( \cos A \) can be derived from the triangle's properties.

### Step 5: Set the Equations Equal
Since \( OD = OI \), we can set the two expressions equal:
\[ r \cos A = \sqrt{R^2 - 2Rr} \]

**Hint:** This step involves equating the two expressions derived from the properties of the square and the triangle.

### Step 6: Square Both Sides
To eliminate the square root, we square both sides:
\[ (r \cos A)^2 = R^2 - 2Rr \]

**Hint:** Be careful with squaring both sides; it can introduce extraneous solutions.

### Step 7: Substitute \( \cos A \)
Using the known relationship \( \cos A = \frac{r}{R} \), substitute this into the equation:
\[ \left(\frac{r}{R}\right)^2 = R^2 - 2Rr \]

**Hint:** This substitution simplifies the equation significantly.

### Step 8: Rearrange the Equation
Rearranging gives:
\[ \frac{r^2}{R^2} = R^2 - 2Rr \]

**Hint:** Make sure to bring all terms to one side to form a standard quadratic equation.

### Step 9: Multiply Through by \( R^2 \)
To eliminate the fraction, multiply through by \( R^2 \):
\[ r^2 = R^4 - 2R^3r \]

**Hint:** This will help in forming a quadratic equation in terms of \( r \).

### Step 10: Rearrange to Standard Form
Rearranging gives:
\[ R^4 - 2R^3r - r^2 = 0 \]

**Hint:** This is now a standard quadratic equation in \( r \).

### Step 11: Solve the Quadratic Equation
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = -1, b = -2R^3, c = R^4 \):
\[ r = \frac{2R^3 \pm \sqrt{(2R^3)^2 - 4(-1)(R^4)}}{2(-1)} \]

**Hint:** Calculate the discriminant carefully.

### Step 12: Simplify the Roots
After simplification, we find:
\[ r = R \sqrt{2 - 1} \]

**Hint:** Ensure to check the conditions under which the roots are valid.

### Final Result
Thus, we conclude that:
\[ \frac{r}{R} = \sqrt{2 - 1} \]

This gives us the relationship between the inradius and circumradius under the given conditions.