Question

If point \(D\) divides base \(BC\) of \(\triangle ABC\) in ratio \(m:n\), then value of \(mBD^2 + nCD^2 + (m+n)AD^2\) is:

• A. \(mAC^2 + nAB^2\)
• B. \((m+n)(AC^2 + AB^2)\)
• C. \(nAC^2 + mAB^2\)
• D. None of these

Hint:

Whenever a point divides a side of a triangle and squares of sides are involved, think of Stewart’s theorem directly.

Answer 1

The Correct Option is C

Concept:
We use Stewart’s Theorem, which applies to a triangle when a point divides one side in a given ratio. For triangle \(ABC\), if point \(D\) lies on \(BC\) such that: \[ BD = \frac{m}{m+n}BC, \quad CD = \frac{n}{m+n}BC \] then Stewart’s theorem states: \[ m \cdot BD^2 + n \cdot CD^2 + (m+n)\cdot AD^2 = n \cdot AC^2 + m \cdot AB^2 \]

Step 1: Identify parameters.
Here, \(D\) divides \(BC\) in the ratio \(m:n\), so: \[ BD : DC = m : n \]

Step 2: Apply Stewart’s theorem.
Using the theorem directly: \[ mBD^2 + nCD^2 + (m+n)AD^2 = nAC^2 + mAB^2 \]

Step 3: Match with given expression.
The expression given in the question is exactly the left-hand side of Stewart’s theorem.

Step 4: Conclusion.
\[ mBD^2 + nCD^2 + (m+n)AD^2 = nAC^2 + mAB^2 \]