Question

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

Hint:

In a right-angled triangle, the length of a median can be calculated using a formula involving the sides of the triangle.

Answer 1

Let \( \triangle ABC \) be a right-angled triangle at \( A \), and let \( BL \) and \( CM \) be the medians from vertices \( B \) and \( C \), respectively. We need to prove that: \[ 4(BL^2 + CM^2) = 5BC^2. \] Step 1: Use the property of medians in a right-angled triangle. In a right-angled triangle, the length of the median from the right angle can be given by the formula: \[ BL^2 = \frac{2AB^2 + 2AC^2 - BC^2}{4}. \] Similarly, the length of the median from vertex \( C \) is: \[ CM^2 = \frac{2BC^2 + 2AC^2 - AB^2}{4}. \] Step 2: Add \( BL^2 \) and \( CM^2 \). Now, adding these two equations for \( BL^2 \) and \( CM^2 \): \[ BL^2 + CM^2 = \frac{2AB^2 + 2AC^2 - BC^2}{4} + \frac{2BC^2 + 2AC^2 - AB^2}{4}. \] Simplifying the right-hand side: \[ BL^2 + CM^2 = \frac{2AB^2 + 2AC^2 - BC^2 + 2BC^2 + 2AC^2 - AB^2}{4}. \] \[ BL^2 + CM^2 = \frac{AB^2 + 4AC^2 + BC^2}{4}. \] Step 3: Multiply by 4 to complete the proof. Now, multiply both sides by 4: \[ 4(BL^2 + CM^2) = AB^2 + 4AC^2 + BC^2. \] Since \( \triangle ABC \) is a right-angled triangle, \( AB^2 + AC^2 = BC^2 \) by the Pythagorean theorem. Thus: \[ 4(BL^2 + CM^2) = BC^2 + 4AC^2 + BC^2 = 5BC^2. \]
Conclusion:
We have shown that \( 4(BL^2 + CM^2) = 5BC^2 \).