Question

ABC is an isosceles right triangle with C as right angle. Prove that \(AB^2 = 2AC^2\).

Hint:

This result is a standard property of isosceles right triangles (also known as 45-45-90 triangles). The ratio of the sides is \(x : x : x\sqrt{2}\). Squaring the hypotenuse gives \((x\sqrt{2})^2 = 2x^2\), which is twice the square of a leg.

Answer 1

Step 1: Understanding the Concept:
This is a direct application of the Pythagorean theorem to an isosceles right-angled triangle. We need to use the properties of both types of triangles to establish the relationship.

Step 2: Detailed Explanation:
Given: \(\triangle ABC\) is an isosceles right triangle, with the right angle at C (\(\angle C = 90^\circ\)).
Since the triangle is isosceles and the right angle is at C, the sides adjacent to the right angle must be equal. Therefore, \(AC = BC\).
AB is the hypotenuse.
By 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. \[ AB^2 = AC^2 + BC^2 \] Now, substitute the condition for an isosceles triangle (\(BC = AC\)) into the Pythagorean theorem: \[ AB^2 = AC^2 + (AC)^2 \] \[ AB^2 = 2AC^2 \]

Step 3: Final Answer:
Hence, it is proved that \(AB^2 = 2AC^2\).