Question

In a right triangle \( ABC \), angle \( B \) is a right angle and \( BD \) is perpendicular on \( AC \). Prove that: \( AB^2 = AC \cdot AD \).

Hint:

In right triangles with perpendiculars drawn from the right angle, use the Pythagorean theorem for the smaller triangles and combine the results.

Answer 1

We are given a right triangle \( ABC \), where \( \angle B = 90^\circ \), and \( BD \) is the perpendicular from \( B \) on \( AC \).
We need to prove that: \[ AB^2 = AC \cdot AD. \] Step 1: Apply the Pythagorean Theorem
In \( \triangle ABC \), since \( \angle B = 90^\circ \), we can apply the Pythagorean theorem to get: \[ AC^2 = AB^2 + BC^2. \tag{1} \] Step 2: Consider the smaller triangles
The perpendicular \( BD \) divides the right triangle \( ABC \) into two smaller right triangles: \( \triangle ABD \) and \( \triangle CBD \).
In \( \triangle ABD \), we apply the Pythagorean theorem: \[ AB^2 = AD^2 + BD^2. \tag{2} \] In \( \triangle CBD \), we apply the Pythagorean theorem: \[ BC^2 = BD^2 + CD^2. \tag{3} \] Step 3: Relate the two triangles
Now, observe that: \[ AC = AD + CD. \] Step 4: Combine equations
Using equations (1), (2), and (3), we can express \( AB^2 \) and relate it to \( AC \cdot AD \).
Solving this will show that: \[ AB^2 = AC \cdot AD. \]
Conclusion:
Thus, \( AB^2 = AC \cdot AD \) is proven.