Question

If $BL$ and $CM$ are medians of a triangle $ABC$ right angled at $A$ drawn from $B$ and $C$ meeting $AC$, $AB$ respectively at $L$, $M$, then $(BL^{2}+CM^{2})=$

• A. $2BC^{2}$
• B. $\frac{3}{4}BC^{2}$
• C. $\frac{5}{4}BC^{2}$
• D. $\frac{6}{4}BC^{2}$

Hint:

For a right-angled triangle $ABC$ right-angled at $A$, if medians are drawn from the acute vertices to the opposite sides, then the standard identity \[ BL^2+CM^2=\frac{5}{4}BC^2 \] is obtained directly by applying the Pythagorean theorem to the two smaller right triangles formed by the medians. This is a useful result frequently used in geometry and competitive examinations.

Answer 1

The Correct Option is C

Concept: The problem involves a right-angled triangle and medians drawn from the two acute vertices. The key idea is to express the lengths of the medians using the Pythagorean theorem and then simplify the resulting expression using the fundamental relation of a right-angled triangle. For a triangle $ABC$ right-angled at $A$: \[ BC^2 = AB^2 + AC^2 \] where $BC$ is the hypotenuse. Also, since a median joins a vertex to the midpoint of the opposite side: \[ AL=\frac{AC}{2} \] and \[ AM=\frac{AB}{2}. \] These relations allow us to express the lengths of the medians $BL$ and $CM$ in terms of the sides $AB$ and $AC$.

Step 1: Write the lengths of the median segments created on the opposite sides. Since $BL$ is the median drawn from vertex $B$ to side $AC$, point $L$ is the midpoint of $AC$. Therefore, \[ AL=LC=\frac{AC}{2}. \] Similarly, $CM$ is the median drawn from vertex $C$ to side $AB$, so point $M$ is the midpoint of $AB$. Hence, \[ AM=MB=\frac{AB}{2}. \] These midpoint relations will be used while applying the Pythagorean theorem.

Step 2: Find an expression for $BL^2$. Consider the right-angled triangle $\triangle ABL$. Since $\angle BAL=90^\circ$, by the Pythagorean theorem, \[ BL^2 = AB^2 + AL^2. \] Substituting \[ AL=\frac{AC}{2}, \] we obtain \[ BL^2 = AB^2+\left(\frac{AC}{2}\right)^2. \] Therefore, \[ BL^2 = AB^2+\frac{AC^2}{4}. \] This is our first required relation. \[ \boxed{BL^2=AB^2+\frac{AC^2}{4}} \]

Step 3: Find an expression for $CM^2$. Now consider the right-angled triangle $\triangle ACM$. Since $\angle CAM=90^\circ$, applying the Pythagorean theorem gives \[ CM^2=AC^2+AM^2. \] Substituting \[ AM=\frac{AB}{2}, \] we get \[ CM^2 = AC^2+\left(\frac{AB}{2}\right)^2. \] Hence, \[ CM^2 = AC^2+\frac{AB^2}{4}. \] Thus, \[ \boxed{CM^2=AC^2+\frac{AB^2}{4}} \]

Step 4: Add the two expressions. We are required to find \[ BL^2+CM^2. \] Adding the expressions obtained above: \[ BL^2+CM^2 = \left(AB^2+\frac{AC^2}{4}\right) + \left(AC^2+\frac{AB^2}{4}\right). \] Grouping like terms, \[ BL^2+CM^2 = \left(AB^2+\frac{AB^2}{4}\right) + \left(AC^2+\frac{AC^2}{4}\right). \] Taking common factors, \[ BL^2+CM^2 = \frac{5AB^2}{4} + \frac{5AC^2}{4}. \] Factoring out $\frac{5}{4}$, \[ BL^2+CM^2 = \frac{5}{4}(AB^2+AC^2). \]

Step 5: Use the Pythagorean theorem in the original triangle. Since $\triangle ABC$ is right-angled at $A$, \[ AB^2+AC^2=BC^2. \] Substituting into the previous result, \[ BL^2+CM^2 = \frac{5}{4}BC^2. \] Therefore, \[ \boxed{BL^2+CM^2=\frac{5}{4}BC^2} \] which matches option (C).