Answer :
To prove the identity [tex]\(\frac{3}{4}(a^2 + b^2 + c^2) = m_a^2 + m_b^2 + m_c^2\)[/tex], where [tex]\(m_a\)[/tex], [tex]\(m_b\)[/tex], and [tex]\(m_c\)[/tex] are the medians of a triangle corresponding to sides [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex], respectively, we can proceed by using the geometric properties of medians in a triangle. However, proving the statement generally involves understanding and applying some advanced geometrical properties and theorems. Below is a step-by-step outline considering such properties:
1. Understanding the Medians of a Triangle:
- A median of a triangle is a line segment drawn from a vertex to the midpoint of the opposite side.
- [tex]\(m_a\)[/tex] denotes the median from the vertex opposite side [tex]\(a\)[/tex], [tex]\(m_b\)[/tex] denotes the median from the vertex opposite side [tex]\(b\)[/tex], and [tex]\(m_c\)[/tex] denotes the median from the vertex opposite side [tex]\(c\)[/tex].
2. Formula for the Length of a Median:
The length of a median in a triangle can be expressed using the following formula:
[tex]\[ m_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2} \][/tex]
Similarly, for the other medians:
[tex]\[ m_b = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2} \][/tex]
[tex]\[ m_c = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2} \][/tex]
3. Expressing the Squares of Medians:
By squaring these formulas, we can express the squares of medians as:
[tex]\[ m_a^2 = \left(\frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}\right)^2 = \frac{1}{4}(2b^2 + 2c^2 - a^2) \][/tex]
[tex]\[ m_b^2 = \frac{1}{4}(2a^2 + 2c^2 - b^2) \][/tex]
[tex]\[ m_c^2 = \frac{1}{4}(2a^2 + 2b^2 - c^2) \][/tex]
4. Summing the Squares of Medians:
Adding the expressions for [tex]\(m_a^2\)[/tex], [tex]\(m_b^2\)[/tex], and [tex]\(m_c^2\)[/tex]:
[tex]\[ m_a^2 + m_b^2 + m_c^2 = \frac{1}{4}(2b^2 + 2c^2 - a^2) + \frac{1}{4}(2a^2 + 2c^2 - b^2) + \frac{1}{4}(2a^2 + 2b^2 - c^2) \][/tex]
Simplifying the right-hand side:
[tex]\[ m_a^2 + m_b^2 + m_c^2 = \frac{1}{4}(2b^2 + 2c^2 - a^2 + 2a^2 + 2c^2 - b^2 + 2a^2 + 2b^2 - c^2) \][/tex]
5. Combining Like Terms:
[tex]\[ m_a^2 + m_b^2 + m_c^2 = \frac{1}{4}(2b^2 + 2c^2 - a^2 + 2a^2 + 2c^2 - b^2 + 2a^2 + 2b^2 - c^2) \][/tex]
Combine the like terms inside the parentheses:
[tex]\[ = \frac{1}{4}(2a^2 + 2a^2 + 2b^2 + 2b^2 + 2c^2 + 2c^2 - a^2 - b^2 - c^2) \][/tex]
[tex]\[ = \frac{1}{4}(4a^2 + 4b^2 + 4c^2 - a^2 - b^2 - c^2) \][/tex]
[tex]\[ = \frac{1}{4}(3a^2 + 3b^2 + 3c^2) \][/tex]
[tex]\[ = \frac{3}{4}(a^2 + b^2 + c^2) \][/tex]
6. Conclusion:
Therefore, we have shown that:
[tex]\[ \frac{3}{4}(a^2 + b^2 + c^2) = m_a^2 + m_b^2 + m_c^2 \][/tex]
as required. This completes the proof.
1. Understanding the Medians of a Triangle:
- A median of a triangle is a line segment drawn from a vertex to the midpoint of the opposite side.
- [tex]\(m_a\)[/tex] denotes the median from the vertex opposite side [tex]\(a\)[/tex], [tex]\(m_b\)[/tex] denotes the median from the vertex opposite side [tex]\(b\)[/tex], and [tex]\(m_c\)[/tex] denotes the median from the vertex opposite side [tex]\(c\)[/tex].
2. Formula for the Length of a Median:
The length of a median in a triangle can be expressed using the following formula:
[tex]\[ m_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2} \][/tex]
Similarly, for the other medians:
[tex]\[ m_b = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2} \][/tex]
[tex]\[ m_c = \frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2} \][/tex]
3. Expressing the Squares of Medians:
By squaring these formulas, we can express the squares of medians as:
[tex]\[ m_a^2 = \left(\frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}\right)^2 = \frac{1}{4}(2b^2 + 2c^2 - a^2) \][/tex]
[tex]\[ m_b^2 = \frac{1}{4}(2a^2 + 2c^2 - b^2) \][/tex]
[tex]\[ m_c^2 = \frac{1}{4}(2a^2 + 2b^2 - c^2) \][/tex]
4. Summing the Squares of Medians:
Adding the expressions for [tex]\(m_a^2\)[/tex], [tex]\(m_b^2\)[/tex], and [tex]\(m_c^2\)[/tex]:
[tex]\[ m_a^2 + m_b^2 + m_c^2 = \frac{1}{4}(2b^2 + 2c^2 - a^2) + \frac{1}{4}(2a^2 + 2c^2 - b^2) + \frac{1}{4}(2a^2 + 2b^2 - c^2) \][/tex]
Simplifying the right-hand side:
[tex]\[ m_a^2 + m_b^2 + m_c^2 = \frac{1}{4}(2b^2 + 2c^2 - a^2 + 2a^2 + 2c^2 - b^2 + 2a^2 + 2b^2 - c^2) \][/tex]
5. Combining Like Terms:
[tex]\[ m_a^2 + m_b^2 + m_c^2 = \frac{1}{4}(2b^2 + 2c^2 - a^2 + 2a^2 + 2c^2 - b^2 + 2a^2 + 2b^2 - c^2) \][/tex]
Combine the like terms inside the parentheses:
[tex]\[ = \frac{1}{4}(2a^2 + 2a^2 + 2b^2 + 2b^2 + 2c^2 + 2c^2 - a^2 - b^2 - c^2) \][/tex]
[tex]\[ = \frac{1}{4}(4a^2 + 4b^2 + 4c^2 - a^2 - b^2 - c^2) \][/tex]
[tex]\[ = \frac{1}{4}(3a^2 + 3b^2 + 3c^2) \][/tex]
[tex]\[ = \frac{3}{4}(a^2 + b^2 + c^2) \][/tex]
6. Conclusion:
Therefore, we have shown that:
[tex]\[ \frac{3}{4}(a^2 + b^2 + c^2) = m_a^2 + m_b^2 + m_c^2 \][/tex]
as required. This completes the proof.
