In ΔABC, DE ∥ BC with D on AB and E on AC. If AB = 15 cm, AC = 20 cm and AE = 12 cm, find DB.

**Theorem**: In a right triangle with legs a, $b$ and hypotenuse $c:a^2+b^2=c^2$.

**Proof using similar triangles**:
In Δ$ABC$ right-angled at $C$, drop perpendicular from $C$ to $AB$ meeting it at $D$.
- Δ$ACD$ ~ Δ$ABC(AA$: shared ∠A, both right-angled).
- Δ$BCD$ ~ Δ$ABC(AA$: shared ∠$B$, both right-angled).
From Δ$ACD$ ~ Δ$ABC:A{C}^2=AD\times AB$.
From Δ$BCD$ ~ Δ$ABC:B{C}^2=BD\times AB$.
Add: $A{C}^2+B{C}^2=AB(AD+BD)=AB\cdot AB=A$$B^2$.
So $a^2+b^2=c^2$. ∎

**Converse**: if $a^2+b^2=c^2$ in a triangle, then the triangle is right-angled at the vertex opposite $c$.

**Pythagorean triples**: integer triples (a, $b$, c) with $a^2+b^2=c^2$. Examples: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25).