Sure, let's go through a detailed step-by-step proof of the given identity:
To prove:
[tex]\[
\frac{3}{4}\left(a^2+b^2+c^2\right) = m_a^2 + m_b^2 + m_c^2
\][/tex]
where [tex]\(m_a\)[/tex], [tex]\(m_b\)[/tex], and [tex]\(m_c\)[/tex] are the lengths of the medians of the triangle.
### Step 1: Understanding the Median Formula
The length of a median in a triangle can be given by:
[tex]\[
m_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}
\][/tex]
Similarly,
[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]
### Step 2: Squaring the Medians
To find [tex]\(m_a^2\)[/tex], [tex]\(m_b^2\)[/tex], and [tex]\(m_c^2\)[/tex]:
[tex]\[
m_a^2 = \left(\frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}\right)^2
= \frac{1}{4} (2b^2 + 2c^2 - a^2)
= \frac{1}{2}(b^2 + c^2) - \frac{1}{4} a^2
\][/tex]
Similarly,
[tex]\[
m_b^2 = \left(\frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2}\right)^2
= \frac{1}{4} (2a^2 + 2c^2 - b^2)
= \frac{1}{2}(a^2 + c^2) - \frac{1}{4} b^2
\][/tex]
[tex]\[
m_c^2 = \left(\frac{1}{2} \sqrt{2a^2 + 2b^2 - c^2}\right)^2
= \frac{1}{4} (2a^2 + 2b^2 - c^2)
= \frac{1}{2}(a^2 + b^2) - \frac{1}{4} c^2
\][/tex]
### Step 3: Adding the Squared Medians
We add [tex]\(m_a^2\)[/tex], [tex]\(m_b^2\)[/tex], and [tex]\(m_c^2\)[/tex]:
[tex]\[
m_a^2 + m_b^2 + m_c^2 = \left(\frac{1}{2}(b^2 + c^2) - \frac{1}{4} a^2 \right) + \left(\frac{1}{2}(a^2 + c^2) - \frac{1}{4} b^2 \right) + \left(\frac{1}{2}(a^2 + b^2) - \frac{1}{4} c^2 \right)
\][/tex]
Combine the terms:
[tex]\[
m_a^2 + m_b^2 + m_c^2 = \left(\frac{1}{2}b^2 + \frac{1}{2}c^2 - \frac{1}{4}a^2\right) + \left(\frac{1}{2}a^2 + \frac{1}{2}c^2 - \frac{1}{4}b^2\right) + \left(\frac{1}{2}a^2 + \frac{1}{2}b^2 - \frac{1}{4}c^2\right)
\][/tex]
Combine similar terms:
[tex]\[
= \frac{1}{2}a^2 + \frac{1}{2}a^2 + \frac{1}{2}b^2 + \frac{1}{2}b^2 + \frac{1}{2}c^2 + \frac{1}{2}c^2 - \frac{1}{4}a^2 - \frac{1}{4}b^2 - \frac{1}{4}c^2
\][/tex]
[tex]\[
= a^2 + b^2 + c^2 - \frac{1}{4}(a^2 + b^2 + c^2)
\][/tex]
[tex]\[
= a^2 + b^2 + c^2 - \frac{1}{4}a^2 - \frac{1}{4}b^2 - \frac{1}{4}c^2
\][/tex]
[tex]\[
= \left(a^2 - \frac{1}{4}a^2\right) + \left(b^2 - \frac{1}{4}b^2\right) + \left(c^2 - \frac{1}{4}c^2\right)
\][/tex]
[tex]\[
= \frac{3}{4}a^2 + \frac{3}{4}b^2 + \frac{3}{4}c^2
\][/tex]
[tex]\[
= \frac{3}{4}(a^2 + b^2 + c^2)
\][/tex]
Thus, we have shown that:
[tex]\[
\frac{3}{4}(a^2 + b^2 + c^2) = m_a^2 + m_b^2 + m_c^2
\][/tex]
This completes the proof.
