Given the condition that the ratio of the diagonal of a rectangle to its length is the same as the ratio of the length of the rectangle to its breadth, and both are equal to $m$, we need to find the value of $m$.

Let the length of the rectangle be $l$, breadth be $b$, and the diagonal be $d$. The condition is given as:

$\frac{d}{l} = \frac{l}{b} = m$

From this, we have two equations:

1. $\frac{d}{l} = m$, which gives $d = m \cdot l$
2. $\frac{l}{b} = m$, which gives $b = \frac{l}{m}$

Next, we apply the Pythagorean theorem to the rectangle, which states that the diagonal $d$ is related to the length and breadth by:

$d^2 = l^2 + b^2$

Substituting $d = m \cdot l$ and $b = \frac{l}{m}$ into this equation, we get:

$(m \cdot l)^2 = l^2 + \left( \frac{l}{m} \right)^2$

Simplifying:

$m^2 l^2 = l^2 + \frac{l^2}{m^2}$

Divide the entire equation by $l^2$ (assuming $l \neq 0$):

$m^2 = 1 + \frac{1}{m^2}$

Multiply both sides by $m^2$ to eliminate the fraction:

$m^4 = m^2 + 1$

Rearrange the equation:

$m^4 - m^2 - 1 = 0$

Let $x = m^2$, so the equation becomes:

$x^2 - x - 1 = 0$

This is a quadratic equation. Solving it using the quadratic formula:

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)} = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}$

Thus:

$x = \frac{1 + \sqrt{5}}{2} \quad \text{or} \quad x = \frac{1 - \sqrt{5}}{2}$

Since $x = m^2$ must be positive, we take the positive root:

$x = \frac{1 + \sqrt{5}}{2}$

Therefore, $m^2 = \frac{1 + \sqrt{5}}{2}$, and taking the square root:

$m = \sqrt{\frac{1 + \sqrt{5}}{2}}$

This is the value of $m$.

Would you like further details or clarifications on any part of this solution?

Here are five related questions:

1. What is the geometric interpretation of the golden ratio, which relates to this solution?
2. How is the Pythagorean theorem applied in different shapes besides rectangles?
3. What are other mathematical problems involving similar ratios in geometry?
4. How can quadratic equations like this be solved graphically?
5. How does this problem relate to the Fibonacci sequence?

Tip: When solving ratios involving geometry, always look for patterns that might simplify through the Pythagorean theorem or algebraic manipulation.