ABCD is a trapezium, AB and CD are parallel sides, M is the midpoint of AD. Prove that the area of the triangle BMC is half the area of the trapezium.
I let A be the origin, then the vectors of AB, AC, and AD respectively are b, c, d. So vector AM=
Area of the trapezium=
vector CM= vector CB=
My problem is I am stuck here, I think it is just some rearrangement of the vectors, but I am no good at dot and cross-products.