The problem asks us to prove that:

$AP \cdot PC = BP \cdot PD$

Given:

• Triangles $\triangle ABC$ and $\triangle DBC$ are right-angled on the same hypotenuse $BC$.
• Line segment $BD$ intersects $AC$ at point $P$.

To Prove: $AP \cdot PC = BP \cdot PD$

This kind of geometric problem usually suggests that we use properties of similar triangles and intersecting chords. Here's how we can prove this:

Proof:

1. Triangles $\triangle ABC$ and $\triangle DBC$: Since both triangles are right-angled on the same hypotenuse $BC$, they share some angles:

   • $\angle ABC = \angle DBC = 90^\circ$
   • $\angle BAC = \angle BDC$ (since they are opposite angles of the right-angled triangles)

2. Use of Intersecting Chord Theorem: The key theorem that applies here is the Intersecting Chord Theorem. This theorem states that if two chords of a circle intersect each other, the product of the segments of one chord equals the product of the segments of the other chord.

   Here, points $A$, $B$, $C$, $D$, and $P$ form such a configuration, with $BD$ intersecting $AC$ at point $P$.

   Therefore, by the intersecting chord theorem, we can directly conclude: $AP \cdot PC = BP \cdot PD$

This concludes the proof.

Would you like to see more details or have any questions?

Here are some related questions to consider:

1. What other theorems in geometry are useful for proving properties of intersecting chords?
2. Can the intersecting chord theorem be used in non-right-angled triangles?
3. How does the similarity of triangles help in geometric proofs?
4. What happens if point $P$ lies on the hypotenuse instead of intersecting $AC$?
5. How would you prove this statement using coordinate geometry?

Tip: When dealing with geometric proofs, drawing diagrams and marking known angles and lengths can often simplify the problem significantly.