ABCD is a trapezium with AB parallel to DC and DC = 3AB. M is the midpoint of DC. `bar"AB" = bar"p", bar"BC" = bar"q"`.

Find in terms of `bar"p" and bar"q"`:

(i) `bar"AM"` (ii) `bar"BD"` (iii) `bar "MB"` (iv) `bar"DA"`

#### Solution

DC is parallel to AB and DC = 3AB.

∵ `bar"AB" = bar"p" therefore bar"DC" = 3bar"p"`

M is the midpoint of DC

∴ `bar"DM" = bar"MC" = 1/2bar"DC" = 3/2bar"p"`

(i) `bar"AM" = bar"AM" + bar"BC" + bar"CM"`

`= bar"AB" + bar"BC" - bar"MC"`

`= bar"p" + bar"q" - "3p"/2`

`= bar"q" - 1/2 bar"p"`

(ii) `bar"BD" = bar"BC" + bar"CD" = bar"BC" - bar"DC" = bar"q" - 3bar"p"`

(iii) `bar"MB" = bar"MC" + bar"CB" = bar"MC" - bar"BC" = 3/2 bar"p" - bar"q"`

(iv) `bar"DA" = bar"DC" + bar"CB" + bar"BA" = bar"DC" - bar"BC" - bar"AB"`

`= 3bar"p" - bar"q" - bar"p" = 2bar"p" - bar"q"`

#### Notes

[**Note:** In the textbook answer instead of p and q, a and b are written.]