To prove that $\Delta ABC$ is a right-angled triangle given that $AD \perp BC$ and $AD^2 = BD \times CD$, we can condense the proof into two key steps:

Step 1: Use the Geometric Mean Theorem

The Geometric Mean Theorem states that if an altitude is drawn from the right angle of a right triangle to the hypotenuse, then the length of the altitude is the geometric mean of the segments of the hypotenuse. For triangle $\Delta ABC$, since $AD \perp BC$, we can apply this theorem:

$AD^2 = BD \times CD$

This is exactly the condition given in the problem. Therefore, this implies that $\Delta ABC$ must be a right-angled triangle, with the right angle at $A$.

Step 2: Apply the Converse of the Pythagorean Theorem

Since $AD^2 = BD \times CD$ is satisfied, and it follows from the Geometric Mean Theorem, $\angle BAC$ must be $90^\circ$. Hence, $\Delta ABC$ is a right-angled triangle by the converse of the Pythagorean Theorem.

Conclusion:

Thus, $\Delta ABC$ is a right-angled triangle with the right angle at $\angle BAC$.

Would you like further clarification or details on any part of the proof?

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Here are five related questions:

1. How can the Converse of the Pythagorean Theorem be used in other geometric proofs?
2. What is the importance of the altitude in right triangles?
3. Can the Geometric Mean Theorem be applied to non-right triangles?
4. How can the property $AD^2 = BD \times CD$ be used to find the length of $AD$ given $BD$ and $CD$?
5. What are the implications if $AD^2 \neq BD \times CD$?

Tip: When proving a triangle is right-angled, check if any geometric mean relationships exist between segments and altitudes.