Question

If D and E are the midpoints of the sides BA and BC of triangle ABC, then \(\overline{AE}+\overline{DC} =\)

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

Hint:

Use position vectors with midpoints, then add AE and DC. The b terms cancel and 3/2 (c - a) remains.

Answer 1

The Correct Option is C

Step 1: Vector chain method:
Take B as origin, with $\overline{BA}=\vec p$ and $\overline{BC}=\vec q$. Then $\overline{BD}=\frac{\vec p}{2}$ and $\overline{BE}=\frac{\vec q}{2}$.

Step 2: Write the two vectors:
$\overline{AE} = \overline{BE}-\overline{BA} = \frac{\vec q}{2}-\vec p$ and $\overline{DC} = \overline{BC}-\overline{BD} = \vec q-\frac{\vec p}{2}$.

Step 3: Add:
$\overline{AE}+\overline{DC} = \frac32\vec q - \frac32\vec p = \frac32(\vec q-\vec p)$.

Step 4: Identify:
$\vec q-\vec p = \overline{BC}-\overline{BA} = \overline{AC}$. So the sum is $\frac32\overline{AC}$, option (C).

Final Answer:
Option (C). \[ \boxed{\frac{3}{2}\overline{AC}} \]