Pythagoras Theorem and Its Proof Using Similarity for CBSE Class 10

Among all the mathematical theorems discovered in human history, none is more universally recognized or widely applied than the Pythagoras Theorem. From ancient Greek geometry and Indian Vedic astronomy (Sulba Sutras) to modern GPS navigation and architectural surveying, this theorem serves as the bedrock of Euclidean distance measurement.

While you are familiar with the formula $a^2 + b^2 = c^2$, CBSE Class 10 Mathematics approaches the theorem from a deeper theoretical foundation: proving the Pythagoras Theorem rigorously using the properties of similar triangles.

What You Will Learn

• Statement of the Pythagoras Theorem
• The foundational Right Triangle Altitude Theorem (NCERT Theorem 6.7)
• Complete, step-by-step geometric proof of Pythagoras Theorem using similarity
• Why similarity provides the cleanest proof of the theorem
• High-yield board exam riders (e.g., median formulas in right-angled triangles)
• Presentation guidelines for full marks in 5-mark Section D questions

1. Statement of the Pythagoras Theorem

Theorem Statement (CBSE Theorem 6.8)

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

$\mathbf{AC^2 = AB^2 + BC^2} \quad (\text{where } \angle B = 90^\circ)$

Here:

• $AC$ is the hypotenuse (the side opposite the right angle, which is always the longest side).
• $AB$ and $BC$ are the other two perpendicular sides (legs).

2. The Perpendicular Altitude Theorem (NCERT Theorem 6.7)

Before proving the Pythagoras Theorem, we need an essential lemma regarding right-angled triangles:

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

                                      B (90°)
                                     /|                                     / |                                     /  |                                     /   |                                     A----D-----C

In right-angled triangle $\Delta ABC$ with $\angle B = 90^\circ$ and $BD \perp AC$:

1. $\mathbf{\Delta ADB \sim \Delta ABC}$
2. $\mathbf{\Delta BDC \sim \Delta ABC}$
3. $\mathbf{\Delta ADB \sim \Delta BDC}$

Quick Verification:

• In $\Delta ADB$ and $\Delta ABC$: $\angle A = \angle A$ (common), $\angle ADB = \angle ABC = 90^\circ$. By AA criterion, $\Delta ADB \sim \Delta ABC$.
• In $\Delta BDC$ and $\Delta ABC$: $\angle C = \angle C$ (common), $\angle BDC = \angle ABC = 90^\circ$. By AA criterion, $\Delta BDC \sim \Delta ABC$.

3. Geometric Proof of the Pythagoras Theorem

Given:

A right-angled triangle $ABC$, right-angled at $B$ ($\angle B = 90^\circ$).

To Prove:

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

Construction:

Draw $BD \perp AC$.

Step-by-Step Proof:

1. Compare Small Left Triangle with the Whole Triangle: In $\Delta ADB$ and $\Delta ABC$:

   • $\angle A = \angle A$ (Common angle)
   • $\angle ADB = \angle ABC = 90^\circ$ (Right angles) Therefore, by the AA Similarity Criterion: $\Delta ADB \sim \Delta ABC$

2. Equate Ratios of Corresponding Sides: $\frac{AD}{AB} = \frac{AB}{AC}$ Cross-multiplying gives: $\mathbf{AB^2 = AD \times AC} \quad \text{--- (1)}$

3. Compare Small Right Triangle with the Whole Triangle: In $\Delta BDC$ and $\Delta ABC$:

   • $\angle C = \angle C$ (Common angle)
   • $\angle BDC = \angle ABC = 90^\circ$ (Right angles) Therefore, by the AA Similarity Criterion: $\Delta BDC \sim \Delta ABC$

4. Equate Ratios of Corresponding Sides: $\frac{CD}{BC} = \frac{BC}{AC}$ Cross-multiplying gives: $\mathbf{BC^2 = CD \times AC} \quad \text{--- (2)}$

5. Add Equation (1) and Equation (2): $AB^2 + BC^2 = (AD \times AC) + (CD \times AC)$

6. Factor out the common term $AC$: $AB^2 + BC^2 = AC \times (AD + CD)$

7. Notice from the geometry diagram: Point $D$ lies on line segment $AC$. Therefore: $AD + CD = AC$

8. Substitute $AC$ for $(AD + CD)$: $AB^2 + BC^2 = AC \times AC$ $\mathbf{AC^2 = AB^2 + BC^2}$ Hence, proved.

Important: <u>This 8-step proof using similarity is one of the standard 5-mark theorem proofs in the CBSE Class 10 board examination. Writing each step with its geometrical justification guarantees full marks.</u>

4. Solved CBSE Board Examination Problems

Solved Example: The Median Theorem Rider (CBSE High-Yield)

Problem: $BL$ and $CM$ are medians of a triangle $ABC$ right-angled at $A$. Prove that $4(BL^2 + CM^2) = 5BC^2$.

Solution:

1. Analyze the Right Triangle at $A$ ($\angle A = 90^\circ$):
   • By Pythagoras theorem in $\Delta ABC$: $BC^2 = AB^2 + AC^2 \quad \text{--- (1)}$
2. In Right Triangle $\Delta ABL$: $BL$ is the hypotenuse: $BL^2 = AL^2 + AB^2$ Since $BL$ is a median, $L$ is the midpoint of $AC$, so $AL = \frac{AC}{2}$: $BL^2 = \left(\frac{AC}{2}\right)^2 + AB^2 = \frac{AC^2}{4} + AB^2$ Multiply by 4: $4BL^2 = AC^2 + 4AB^2 \quad \text{--- (2)}$
3. In Right Triangle $\Delta CMA$: $CM$ is the hypotenuse: $CM^2 = AM^2 + AC^2$ Since $CM$ is a median, $M$ is the midpoint of $AB$, so $AM = \frac{AB}{2}$: $CM^2 = \left(\frac{AB}{2}\right)^2 + AC^2 = \frac{AB^2}{4} + AC^2$ Multiply by 4: $4CM^2 = AB^2 + 4AC^2 \quad \text{--- (3)}$
4. Add Equation (2) and Equation (3): $4BL^2 + 4CM^2 = (AC^2 + 4AB^2) + (AB^2 + 4AC^2)$ $4(BL^2 + CM^2) = 5AB^2 + 5AC^2 = 5(AB^2 + AC^2)$
5. Substitute Equation (1) ($AB^2 + AC^2 = BC^2$): $\mathbf{4(BL^2 + CM^2) = 5BC^2}$ Hence, proved.

5. Summary and Examination Tips

Exam Tip: In the construction, clearly state "Draw $BD \perp AC$". Drawing this single perpendicular line splits the main triangle into two similar sub-triangles, which is the entire engine of the proof!

Common Mistake: Confusing vertex correspondence when equating sides of similar triangles. In $\Delta ADB \sim \Delta ABC$, hypotenuse $AB$ corresponds to hypotenuse $AC$, while base $AD$ corresponds to base $AB$.