Introduction:

In this proof, we will demonstrate that the area of a right-angled triangle with a given hypotenuse is maximum when the triangle is isosceles. To do this, we will use the concept of area and apply basic geometry principles to derive the result.

Proof:

1. Understanding the Problem:
We are given a right-angled triangle with a fixed hypotenuse length. We need to determine the values of the other two sides (base and height) in order to maximize the area of the triangle.

2. Let's Assume:
Let's assume the hypotenuse has a length of 'c'. We can consider the other two sides as 'x' and 'y', where 'x' is the base and 'y' is the height of the triangle.

3. Applying the Pythagorean Theorem:
According to the Pythagorean theorem, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Using this theorem, we have:
x^2 + y^2 = c^2

4. Expressing 'y' in terms of 'x':
To maximize the area of the triangle, we need to express 'y' in terms of 'x'. Rearranging the equation from step 3, we get:
y^2 = c^2 - x^2
y = sqrt(c^2 - x^2)

5. Deriving the Area:
The area of a triangle can be calculated using the formula: A = (1/2) * base * height. Substituting the values of 'x' and 'y' into this formula, we get:
A = (1/2) * x * sqrt(c^2 - x^2)

6. Differentiating the Area:
To find the maximum area, we take the derivative of the area function with respect to 'x' and equate it to zero.
dA/dx = (1/2) * (sqrt(c^2 - x^2) - (x^2 / sqrt(c^2 - x^2))) = 0

7. Solving for 'x':
Simplifying the equation from step 6, we get:
sqrt(c^2 - x^2) - (x^2 / sqrt(c^2 - x^2)) = 0
sqrt(c^2 - x^2) = (x^2 / sqrt(c^2 - x^2))
c^2 - x^2 = x^2
2x^2 = c^2
x^2 = c^2 / 2
x = c / sqrt(2)

8. Finding 'y' and the Maximum Area:
Substituting the value of 'x' into the equation from step 4, we get:
y = sqrt(c^2 - (c / sqrt(2))^2)
y = sqrt(c^2 - c^2 / 2)
y = sqrt(c^2 / 2)
y = c / sqrt(2)

Therefore, the base and height of the triangle are x = c / sqrt(2) and y = c / sqrt(2), respectively