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**arze** 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=$\displaystyle \frac{1}{2}d$

Area of the trapezium=$\displaystyle \frac{1}{2}|b\times c|+\frac{1}{2}|c\times d|$

vector CM=$\displaystyle \frac{1}{2}d-c$ vector CB=$\displaystyle b-c$

Area BMC=$\displaystyle \frac{1}{2}|(b-c)\times (\frac{1}{2}d-c)|$

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.

Thanks