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

Step 1: Understanding the Question:
The question asks for two parts: first, to state the Pythagoras Theorem, and second, to provide a formal geometric proof of the theorem.
Step 2: Key Formula or Approach:
The proof utilizes the concept of similar triangles. By constructing an altitude from the right angle to the hypotenuse, we create two smaller triangles that are similar to the original triangle and to each other. We can then use the property of similar triangles that the ratio of their corresponding sides is equal.
Step 3: Detailed Explanation (Statement and Proof):
Statement:
In a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides (the legs).
Proof:
Given: A triangle \(\triangle ABC\) which is right-angled at B.
To Prove: \(AC^2 = AB^2 + BC^2\).
Proof Steps:
1. Consider the triangles \(\triangle ADB\) and \(\triangle ABC\).
- \(\angle A = \angle A\) (Common angle)
- \(\angle ADB = \angle ABC\) (Both are \(90^\circ\))
- Therefore, by the Angle-Angle (AA) similarity criterion, \(\triangle ADB \sim \triangle ABC\).
2. Since the triangles are similar, the ratio of their corresponding sides is equal:
\[ \frac{AD}{AB} = \frac{AB}{AC} \] Cross-multiplying gives:
\[ AB^2 = AD \cdot AC \quad \cdots (1) \] 3. Now, consider the triangles \(\triangle BDC\) and \(\triangle ABC\).
- \(\angle C = \angle C\) (Common angle)
- \(\angle BDC = \angle ABC\) (Both are \(90^\circ\))
- Therefore, by the AA similarity criterion, \(\triangle BDC \sim \triangle ABC\).
4. The ratio of their corresponding sides is also equal:
\[ \frac{CD}{BC} = \frac{BC}{AC} \] Cross-multiplying gives:
\[ BC^2 = CD \cdot AC \quad \cdots (2) \] 5. Now, add equation (1) and equation (2):
\[ AB^2 + BC^2 = (AD \cdot AC) + (CD \cdot AC) \] 6. Factor out AC from the right-hand side:
\[ AB^2 + BC^2 = AC (AD + CD) \] 7. From the diagram, it is clear that \(AD + CD = AC\). Substitute this into the equation:
\[ AB^2 + BC^2 = AC (AC) \] \[ AB^2 + BC^2 = AC^2 \] This proves the theorem.
Step 4: Final Answer:
Hence, it is proved that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
\[ \boxed{AC^2 = AB^2 + BC^2} \]