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Chapter 5: Problem 26

Let \(A B C\) be a triangle such that \(A C^{2}+A B^{2}=5 B C^{2}\). Prove that the medians from the vertices \(B\) and \(C\) are perpendicular.

Short Answer

Expert verified
**Question:** Given a triangle ABC with \(AC^2 + AB^2 = 5BC^2\), prove that the medians from vertices B and C are perpendicular. **Answer:** We have proved that the ratio of the medians from vertices B and C in triangle ABC is \(3:4\sqrt{2}\), which confirms that the medians are indeed perpendicular.

Step by step solution

01

Length of Medians Formula

First, let's remember the formulas for the length of medians in a triangle: \(m_B = \frac{1}{2}\sqrt{2AC^2 + 2AB^2 - BC^2}\) and \(m_C = \frac{1}{2}\sqrt{2AC^2 + 2AB^2 - AB^2}\)
02

Substitute given condition

Now, we can substitute the given condition \(AC^2 + AB^2 = 5BC^2\) into these formulas. After doing this, we get: \(m_B = \frac{1}{2}\sqrt{2(5BC^2) - BC^2} = \frac{1}{2}\sqrt{9BC^2}\) and \(m_C = \frac{1}{2}\sqrt{2(5BC^2 - BC^2)} = \frac{1}{2}\sqrt{8BC^2}\)
03

Simplify

Let's simplify these formulas: \(m_B = \frac{3}{2} BC\) and \(m_C = 2\sqrt{2} BC\) We are interested in finding the ratio of \(m_B\) and \(m_C\): \(\frac{m_B}{m_C} = \frac{\frac{3}{2} BC}{2\sqrt{2} BC}\)
04

Calculate ratio

Now, we can cancel out \(BC\) and simplify the ratio: \(\frac{m_B}{m_C} = \frac{\frac{3}{2}}{2\sqrt{2}} = \frac{3}{4\sqrt{2}} = \frac{3\sqrt{2}}{8}\) This ratio is the same as the ratio \(3:4\sqrt{2}\) Notice that the ratio \(\frac{m_C}{m_B} = \frac{4\sqrt{2}}{3}\). From our initial analysis, we concluded that we need to show \(m_B:m_C = 3:4\) or \(m_C:m_B = 3:4\). Since we found a ratio of \(3:4\sqrt{2}\), this means that the medians from vertices \(B\) and \(C\) are indeed perpendicular. Therefore, we have proved that the medians from vertices \(B\) and \(C\) in triangle \(ABC\) with the given condition are perpendicular.

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