Prove that the median of a trapezium is parallel to the parallel sides of the trapezium and its length is half of the sum of the lengths of the parallel sides.

#### Solution

Let `bar"a", bar"b", bar"c" "and" bar"d"` be respectively the position vectors of the vertices A, B, C and D of the trapezium ABCD, with side AD || side BC.

Then the vectors `bar"AD"` and `bar"BC"` are parallel.

∴ there exists a scalar k,

such that `bar"AD" = "k".bar"BC"`

∴ `bar"AD" + bar"BC" = "k".bar"BC" + bar"BC"`

`= ("k" + 1)bar"BC"` ....(1)

Let `bar"m"` and `bar"n"` be the position vectors of the midpoints M and N of the non-parallel sides AB and DC respectively. Then seg MN is the median of the trapezium.

By the midpoint formula,

`bar"m" = (bar"a" + bar"b")/2` and `bar"n" = (bar"d" + bar"c")/2`

∴ `bar"MN" = bar"n" - bar"m"`

`= ((bar"d" + bar"c")/2) - ((bar"a" + bar"b")/2)`

`= 1/2 (bar"d" + bar"c" - bar"a" - bar"b")`

`= 1/2 [(bar"d" - bar"a") + (bar"c" - bar"b")]`

`= (bar"AD" + bar"BC")/2` .....(2)

`= (("k" + 1)bar"BC")/2` ....[By (1)]

Thus `bar "MN"` is a scalar multiple of `bar "BC"`

∴ `bar "MN"` and `bar "BC"` are parallel vectors

∴ `bar "MN" || bar "BC"` where `bar "BC" || bar "AD"`

∴ the median MN is parallel to the parallel sides AD and BC of the trapezium.

Now `bar "AD"` and `bar "BC"` are collinear

∴ `|bar "AD" + bar "BC"| = |bar "AD"| + |bar "BC"| = "AD" + "BC"`

∴ from (2), we have

Now `bar "MN" = (bar "AD" + bar "BC")/2`

∴ MN = `1/2("AD" + "BC")`