MP Board · Class 10 · Mathematics · TrianglesState and prove the Pythagoras Theorem (Theorem 6.8 of NCERT Class 10 Mathematics).

Statement of Pythagoras Theorem\nIn a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given Data\nLet $\triangle ABC$ be a right-angled triangle, right-angled at $B$ (i.e., $\angle B = 90^\circ$).

To Prove

$$AC^2 = AB^2 + BC^2$$

Construction\nDraw a perpendicular $BD$ from the vertex $B$ to the hypotenuse $AC$ (where $D$ lies on $AC$).

Proof

1. Compare Smaller Triangle with the Main Triangle: Compare $\triangle ADB$ and $\triangle ABC$:

   • $\angle ADB = \angle ABC = 90^\circ$ (By construction and given data)
   • $\angle DAB = \angle CAB$ (Common angle for both triangles)

   Therefore, by AA similarity criterion: $$\triangle ADB \sim \triangle ABC$$

2. Write Ratio of Corresponding Sides: Since the triangles are similar, their corresponding sides are proportional: $$\frac{AD}{AB} = \frac{AB}{AC}$| Cross-multiplying gives: $$AB^2 = AD \times AC \quad \text{--- (Equation 1)}$|

3. Compare Other Smaller Triangle with the Main Triangle: Now, compare $\triangle BDC$ and $\triangle ABC$:

   • $\angle BDC = \angle ABC = 90^\circ$ (By construction and given data)
   • $\angle DCB = \angle BCA$ (Common angle for both triangles)

   Therefore, by AA similarity criterion: $$\triangle BDC \sim \triangle ABC$$

4. Write Ratio of Corresponding Sides: Proportionality of corresponding sides gives: $$\frac{CD}{BC} = \frac{BC}{AC}$| Cross-multiplying gives: $$BC^2 = CD \times AC \quad \text{--- (Equation 2)}$|

5. Adding Equations 1 and 2: Adding Equation 1 and Equation 2 together: $$AB^2 + BC^2 = (AD \times AC) + (CD \times AC)$| Factor out $AC$ from the right side: $$AB^2 + BC^2 = AC \times (AD + CD)$| From the figure, $AD + CD = AC$. Substituting this: $$AB^2 + BC^2 = AC \times AC$$n $$AB^2 + BC^2 = AC^2$$n
   Therefore, $AC^2 = AB^2 + BC^2$. \nHence proved.