Question

State and prove the Pythagoras Theorem.

Hint:

Pythagoras Theorem applies only to right-angled triangles. Remember the formula: \[ (\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Perpendicular})^2 \] Useful in distance formula and coordinate geometry.

Answer 1

Statement (Pythagoras Theorem): In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. 
Given: In $\triangle ABC$, $\angle B = 90^\circ$. So, $AC$ is the hypotenuse. 
To Prove: \[ AC^2 = AB^2 + BC^2 \] Proof (Using similarity): 
Step 1: Draw altitude Draw $BD \perp AC$ from the right angle $B$ to the hypotenuse $AC$. 
Step 2: Consider similar triangles Triangles $\triangle ABC$, $\triangle ADB$, and $\triangle BDC$ are similar because they share common angles. 
From similarity of $\triangle ABC \sim \triangle ADB$: \[ \frac{AB}{AC} = \frac{AD}{AB} \] \[ AB^2 = AC \cdot AD \quad \cdots (1) \] 
From similarity of $\triangle ABC \sim \triangle BDC$: \[ \frac{BC}{AC} = \frac{DC}{BC} \] \[ BC^2 = AC \cdot DC \quad \cdots (2) \] Step 3: Add equations (1) and (2) \[ AB^2 + BC^2 = AC \cdot AD + AC \cdot DC \] \[ AB^2 + BC^2 = AC(AD + DC) \] But, \[ AD + DC = AC \] So, \[ AB^2 + BC^2 = AC^2 \] Conclusion: Hence proved, \[ \boxed{AC^2 = AB^2 + BC^2} \]