Q69 / 150 · MHT CET 2026 (11 April Shift 1) · Mathematics
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$\dfrac{3}{2} \overline{BC}$
- Correct answerC$\dfrac{3}{2} \overline{AC}$
- D$\dfrac{1}{2} \overline{AC}$
Solution
Using the triangle law of vector addition:
$\overline{AE} = \overline{AB} + \overline{BE}$
$\overline{DC} = \overline{DB} + \overline{BC}$
Since D is the midpoint of BA, $\overline{DB} = \dfrac{1}{2}\overline{AB}$.
Since E is the midpoint of BC, $\overline{BE} = \dfrac{1}{2}\overline{BC}$.
Substituting these into the expressions for $\overline{AE}$ and $\overline{DC}$:
$\overline{AE} + \overline{DC} = \left(\overline{AB} + \dfrac{1}{2}\overline{BC}\right) + \left(\dfrac{1}{2}\overline{AB} + \overline{BC}\right)$
$\overline{AE} + \overline{DC} = \dfrac{3}{2}\overline{AB} + \dfrac{3}{2}\overline{BC}$
$\overline{AE} + \overline{DC} = \dfrac{3}{2}(\overline{AB} + \overline{BC})$
By the triangle law of vector addition, $\overline{AB} + \overline{BC} = \overline{AC}$.
$\overline{AE} + \overline{DC} = \dfrac{3}{2}\overline{AC}$
Answer: $\dfrac{3}{2} \overline{AC}$
$\overline{AE} = \overline{AB} + \overline{BE}$
$\overline{DC} = \overline{DB} + \overline{BC}$
Since D is the midpoint of BA, $\overline{DB} = \dfrac{1}{2}\overline{AB}$.
Since E is the midpoint of BC, $\overline{BE} = \dfrac{1}{2}\overline{BC}$.
Substituting these into the expressions for $\overline{AE}$ and $\overline{DC}$:
$\overline{AE} + \overline{DC} = \left(\overline{AB} + \dfrac{1}{2}\overline{BC}\right) + \left(\dfrac{1}{2}\overline{AB} + \overline{BC}\right)$
$\overline{AE} + \overline{DC} = \dfrac{3}{2}\overline{AB} + \dfrac{3}{2}\overline{BC}$
$\overline{AE} + \overline{DC} = \dfrac{3}{2}(\overline{AB} + \overline{BC})$
By the triangle law of vector addition, $\overline{AB} + \overline{BC} = \overline{AC}$.
$\overline{AE} + \overline{DC} = \dfrac{3}{2}\overline{AC}$
Answer: $\dfrac{3}{2} \overline{AC}$
Hence, the correct answer is option: 'C'.