In a right angled triangle, the square of the hypotenuse is twice the product…

For a right triangle with legs b and p and hypotenuse h, the Pythagorean theorem gives h2 = b2 + p2. For any two lengths, (b − p)2 = b2 + p2 − 2bp, so b2 + p2 ≥ 2bp, with equality only when b = p.

1. The given condition is h2 = 2bp.

2. By the Pythagorean theorem, h2 = b2 + p2, so b2 + p2 = 2bp.

3. Rearranging: b2 − 2bp + p2 = 0, i.e. (b − p)2 = 0.

4. This forces b = p — the two legs are equal, so the triangle is an isosceles right triangle.

5. In this triangle the two acute angles are equal, and together with the 90° right angle they sum to 180°, so each acute angle = (180° − 90°) / 2 = 45°.

Cross-check: with b = p, h2 = b2 + p2 = 2b2 = 2·b·p (since p = b), matching the given condition h2 = 2bp — confirming the result. Equivalently, tan(45°) = opposite/adjacent = p/b = 1, consistent with equal legs.