Hello, wahhdoe!

I'll solve it in my usual baby-talk way . . .

$\displaystyle ABCD\text{ is a trapezium with }BC \parallel AD.$

$\displaystyle \overrightarrow{AB} = 3\vec b,\;\; \overrightarrow{BC} = 3\vec a,\;\; \overrightarrow{AD} = 9\vec a$

$\displaystyle M\text{ is the midpoint of }BC\text{ and }N\text{ is the midpoint of }AD.$

$\displaystyle \text{a) Find the vector }\overrightarrow{MN}\text{ in terms of }\vec a\text{ and }\vec b$

. . $\displaystyle \text{-- which I found to be }3\vec a-3\vec b$ . Yes!

$\displaystyle X\text{ is the midpoint of }MN\text{ and }Y\text{ is the midpoint of }CD.$

$\displaystyle \text{b) Prove that }XY \parallel AD$ Code:

: - - 3a - - :
M
B o * * o * * o C
* * *
* * * Y
3b * X o * * * * * o
* * *
* * *
A o * * * * * o * * * * * o D
N
: - - - - - - - -9a - - - - - - - - :

$\displaystyle (a)\;\overrightarrow{MN} \:=\:\overrightarrow{MB} + \overrightarrow{BA} + \overrightarrow{AN}$

. . . . . . .$\displaystyle =\;\text{-}\tfrac{1}{2}\overrightarrow{BC} - \overrightarrow{AB} + \tfrac{1}{2}\overrightarrow{AD}$

. . . . . . .$\displaystyle =\;\text{-}\tfrac{1}{2}(3\vec a) - (3\vec b) + \tfrac{1}{2}(9\vec a)$

. . . . . . .$\displaystyle =\;\text{-}\tfrac{3}{2}\vec a - 3\vec b + \tfrac{9}{2}\vec a$

. . . . . . .$\displaystyle =\;3\vec a - 3\vec b$

$\displaystyle (b)\;\overrightarrow{CD} \;=\;\overrightarrow{CB} + \overrightarrow{BA} + \overrightarrow{AD}$

. . . . . . $\displaystyle =\;-3\vec a - 3\vec b + 9\vec a$

. . . . . . $\displaystyle =\;6\vec a - 3\vec b$

$\displaystyle \overrightarrow{XY} \;=\;\overrightarrow{XM} + \overrightarrow{MC} + \overrightarrow{CY}$

. . . .$\displaystyle =\;\text{-}\tfrac{1}{2}\overrightarrow{MN} + \tfrac{1}{2}\overrightarrow{BC} + \tfrac{1}{2}\overrightarrow{CD}$

. . . .$\displaystyle =\;\text{-}\tfrac{1}{2}(3\vec a - 3\vec b) + \tfrac{1}{2}(3\vec a) + \tfrac{1}{2}(6\vec a - 3\vec b) $

. . . .$\displaystyle =\;\text{-}\tfrac{3}{2}\vec a + \tfrac{3}{2}\vec b + \tfrac{3}{2}\vec a + 3\vec a - \tfrac{3}{2}\vec b$

. . . .$\displaystyle =\; 3\vec a$

$\displaystyle \text{We have: }\:\overrightarrow{XY} \:=\:3\vec a \:=\:\tfrac{1}{3}\overrightarrow{AD}$

$\displaystyle \text{Therefore: }\:\overrightarrow{XY} \parallel \overrightarrow{AD}$