$B L$ and $C M$ are medians of a triangle $A B C$ right-angled at $A$ . Prove that $4 (B L^{2} + C M^{2}) = 5 B C^{2}$ .

hard 5 marks Class 10 · Mathematics · Triangles

Answer

Hence proved,

4 (B L^{2} + C M^{2}) = 5 B C^{2}

The correct answer is Hence proved, $4 (B L^{2} + C M^{2}) = 5 B C^{2}$ . — Substitute $A B^{2} + A C^{2} = B C^{2}$ from Equation 1 to get $4 (B L^{2} + C M^{2}) = 5 B C^{2}$ .

Step-by-step solution

In $△ A B C$ , $∠ A = 9 0^{\circ}$ . Since $B L$ and $C M$ are medians, $L$ is the midpoint of $A C$ and $M$ is the midpoint of $A B$ . Thus, $A L = \frac{A C}{2}$ and $A M = \frac{A B}{2}$ .
Apply the Pythagoras theorem in the right-angled triangle $A B C$ : $B C^{2} = A B^{2} + A C^{2}$ (Equation 1).
Apply the Pythagoras theorem in right-angled triangle $A B L$ : $B L^{2} = A B^{2} + A L^{2}$ . Substitute $A L = \frac{A C}{2}$ to get $B L^{2} = A B^{2} + (\frac{A C}{2})^{2} = A B^{2} + \frac{A C ^{2}}{4}$ .
Multiply the entire equation by $4$ : $4 B L^{2} = 4 A B^{2} + A C^{2}$ (Equation 2).
Apply the Pythagoras theorem in right-angled triangle $C M A$ : $C M^{2} = A C^{2} + A M^{2}$ . Substitute $A M = \frac{A B}{2}$ to get $C M^{2} = A C^{2} + (\frac{A B}{2})^{2} = A C^{2} + \frac{A B ^{2}}{4}$ .
Multiply the entire equation by $4$ : $4 C M^{2} = 4 A C^{2} + A B^{2}$ (Equation 3).
Add Equation 2 and Equation 3: $4 B L^{2} + 4 C M^{2} = (4 A B^{2} + A C^{2}) + (4 A C^{2} + A B^{2}) = 5 A B^{2} + 5 A C^{2} = 5 (A B^{2} + A C^{2})$ .
Substitute $A B^{2} + A C^{2} = B C^{2}$ from Equation 1 to get $4 (B L^{2} + C M^{2}) = 5 B C^{2}$ .

Common mistake: Forgetting to write

A L

\frac{A C}{2}

and

A M

\frac{A B}{2}

before applying the Pythagoras theorem in the smaller right triangles.

Concept tested: Pythagoras Theorem applied to medians of a right triangle

Published 20 August 2026 · How this solution was prepared and checked

BL and CM are medians of a triangle ABC right-angled at A. Prove that 4(BL2+CM2)=5BC2.

Step-by-step solution

Related questions

$B L$ and $C M$ are medians of a triangle $A B C$ right-angled at $A$ . Prove that $4 (B L^{2} + C M^{2}) = 5 B C^{2}$ .