State and prove Pythagoras' theorem.

Q.State and prove Pythagoras' theorem.

**Pythagoras' Theorem:** The area of the square formed on the hypotenuse of any right-angled triangle is equal to the sum of the areas of the squares formed on the other two sides. **Given:** \( \triangle ABC \) is a right-angled triangle where \( \angle A \) is the right angle. **To Prove:** \[ BC^2 = AB^2 + AC^2 \] **Construction:** From the right-angled vertex \( A \), draw a perpendicular \( AD \) on the hypotenuse \( BC \), meeting \( BC \) at point \( D \). **Proof:** Since \( AD \) is perpendicular to \( BC \) in the right-angled triangle \( \triangle ABC \), \(\triangle ABD \) and \( \triangle CBA \) are similar. \[ \therefore \frac{AB}{BC} = \frac{BD}{AB} \] \[ \therefore AB^2 = BC \cdot BD \quad (i) \] Similarly, \( \triangle CAD \) and \( \triangle CBA \) are similar. \[ \therefore \frac{AC}{BC} = \frac{DC}{AC} \quad (ii) \] Adding equations (i) and (ii), \[ AB^2 + AC^2 = BC \cdot BD + BC \cdot DC \] \[ = BC(BD + DC) \] \[ = BC \cdot BC = BC^2 \] \[ \therefore BC^2 = AB^2 + AC^2 \quad \text{(Proved)} \]