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

To solve this problem, we need to use the concept of mass points or weighted averages in coordinate geometry. In this problem, point \( D \) divides the base \( BC \) of triangle \( \triangle ABC \) in the ratio \( m:n \). We are asked to find the expression for \( mBD^2 + nCD^2 + (m+n)AD^2 \) and match it with one of the given options.

1. We start with the section formula, which states that the coordinates of a point \( D \) dividing a line segment \( BC \) in the ratio \( m:n \) are given by: \(D = \left( \frac{mx_B + nx_C}{m+n}, \frac{my_B + ny_C}{m+n} \right)\) Here, \( (x_B, y_B) \) and \( (x_C, y_C) \) are the coordinates of points \( B \) and \( C \) respectively.
2. To find the expression for the required combinations of squares of the sides, use the distance formula. The square of the distance from point \( D \) to points \( B \), \( C \), and \( A \) are calculated as follows: - \(BD^2 = \left( x_D - x_B \right)^2 + \left( y_D - y_B \right)^2\) - \(CD^2 = \left( x_D - x_C \right)^2 + \left( y_D - y_C \right)^2\) - \(AD^2 = \left( x_D - x_A \right)^2 + \left( y_D - y_A \right)^2\)
3. Substituting the expressions of \( x_D \) and \( y_D \) from the section formula into these equations, we simplify to find the following linear combination: - \(mBD^2 + nCD^2 + (m+n)AD^2 = n(x_C-x_A)^2 + n(y_C-y_A)^2 + m(x_B-x_A)^2 + m(y_B-y_A)^2\) This further simplifies, using the equation for distance square, to: \(= nAC^2 + mAB^2\)
4. Comparing this result with each option given, we see that the correct expression is found in the option that equals \( nAC^2 + mAB^2 \).

Thus, the correct answer is: \( nAC^2 + mAB^2 \).