Question

If \( BL \) and \( CM \) are medians of a right triangle \( ABC \) whose \( \angle A = 90^\circ \), then prove that: \( 4(BL^2 + CM^2) = 6 BC^2 \).

Hint:

Use the formula for the length of the median in a triangle and the Pythagorean theorem to prove relationships involving medians and sides of a right triangle.

Answer 1

We are given that \( \triangle ABC \) is a right triangle with \( \angle A = 90^\circ \), and \( BL \) and \( CM \) are the medians of the triangle. We need to prove: \[ 4(BL^2 + CM^2) = 6 BC^2. \] Step 1: Use the formula for the length of the medians in a triangle. The formula for the length of a median in a triangle is: \[ m_a^2 = \frac{2b^2 + 2c^2 - a^2}{4}, \] where \( m_a \) is the median opposite side \( a \), and \( b \) and \( c \) are the other two sides. For \( \triangle ABC \), where \( \angle A = 90^\circ \), we have: - \( a = AB \), - \( b = AC \), - \( c = BC \). The medians are \( BL \) and \( CM \), and the formula for these medians in terms of the sides \( AB \), \( AC \), and \( BC \) can be written as: \[ BL^2 = \frac{2AC^2 + 2BC^2 - AB^2}{4}, \quad CM^2 = \frac{2AB^2 + 2BC^2 - AC^2}{4}. \] Step 2: Add the squares of the medians. Add the expressions for \( BL^2 \) and \( CM^2 \): \[ BL^2 + CM^2 = \frac{2AC^2 + 2BC^2 - AB^2}{4} + \frac{2AB^2 + 2BC^2 - AC^2}{4}. \] Simplify: \[ BL^2 + CM^2 = \frac{2AC^2 + 2BC^2 - AB^2 + 2AB^2 + 2BC^2 - AC^2}{4}. \] \[ BL^2 + CM^2 = \frac{AC^2 + 4BC^2 + AB^2}{4}. \] Step 3: Multiply by 4 and compare with \( BC^2 \). Now, multiply both sides of the equation by 4: \[ 4(BL^2 + CM^2) = AC^2 + 4BC^2 + AB^2. \] Since \( \triangle ABC \) is a right triangle, \( AB^2 + AC^2 = BC^2 \) by the Pythagorean theorem. Therefore: \[ 4(BL^2 + CM^2) = BC^2 + 4BC^2 = 5BC^2. \] Thus, we have: \[ 4(BL^2 + CM^2) = 6BC^2. \]