Question

In parallelogram \( ABCD \), points \( M \) and \( N \) are midpoints of sides \( BC \) and \( CD \) respectively. Then the length \( AM + AN = \) ?

• A. \( \frac{1}{3} AC \)
• B. \( \frac{2}{3} AC \)
• C. \( \frac{3}{4} AC \)
• D. \( \frac{3}{2} AC \)

Hint:

Midpoints in vector geometry simplify using averages. Combine and simplify using vector operations to relate to diagonals or other sides.

Answer 1

The Correct Option is D

Let’s use vector geometry. Let \( \vec{A} = \vec{0}, \vec{B} = \vec{b}, \vec{D} = \vec{d} \), so \( \vec{C} = \vec{b} + \vec{d} \) Then: \[ \vec{M} = \frac{\vec{B} + \vec{C}}{2} = \frac{\vec{b} + (\vec{b} + \vec{d})}{2} = \frac{2\vec{b} + \vec{d}}{2} \] \[ \vec{N} = \frac{\vec{C} + \vec{D}}{2} = \frac{(\vec{b} + \vec{d}) + \vec{d}}{2} = \frac{\vec{b} + 2\vec{d}}{2} \] Now compute: \[ \vec{AM} + \vec{AN} = \vec{M} + \vec{N} = \frac{2\vec{b} + \vec{d} + \vec{b} + 2\vec{d}}{2} = \frac{3\vec{b} + 3\vec{d}}{2} = \frac{3}{2} (\vec{b} + \vec{d}) = \frac{3}{2} \vec{AC} \]