Question
Using vectors prove that the line segment joining the mid-points of non-parallel sides of a trapezium is parallel to the base and is equal to half the sum of the parallel sides.
✓
Answer
Let $\overrightarrow{a}, \overrightarrow{b} \text{and}\overrightarrow{c}$ be the position vectors of A, B and C respectively w.r.t. O. Let D and E be the mid-points of parallel sides OC and AB respectively.
$\therefore \text{Position vector of}$
$\text{D is} \frac{O +\overrightarrow{C}}{2} = \frac{\overrightarrow{c}}{2}$
$CB\parallel OA \Rightarrow \overrightarrow{a}\parallel \overrightarrow{b} - \overrightarrow{c}$
$\Rightarrow \text{E is parallel to base and its length is half the sum of lengths of the parallel sides.}$
$\therefore \text{Position vector of}$$\text{D is} \frac{O +\overrightarrow{C}}{2} = \frac{\overrightarrow{c}}{2}$
$CB\parallel OA \Rightarrow \overrightarrow{a}\parallel \overrightarrow{b} - \overrightarrow{c}$
$\Rightarrow \text{E is parallel to base and its length is half the sum of lengths of the parallel sides.}$
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