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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
Solution
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
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