Question

ABC is a right triangle with angle B = 90°, A circle with BC as diameter meets hypotenuse AC at point D. Prove that: AC × AD = AB²

Answer 1

In ΔABC,

∠B = 90° and BC is the diameter of the circle.

Therefore, AB is the tangent to the circle at B.

Now, AB is tangent and ADC is the secant

∴ AB² = AD × AC