Tom and James from Queen Mary's Grammar School, Walsall and Shu Cao from Oxford High School gave this solution for a convex quadrilateral. Can you see how to adapt the solution for the case of the arrow shaped quadrilateral?

Suppose there is a convex quadrilateral ABCD, the diagonals AC and BD cross each other at O. The angle between AO and BO is q degrees, the angle between DO and CO is the same. The angle between AO and DO is 180-q degrees, the angle between BO and CO is the same.

The area of triangle AOB is ¹/₂AO×BO sinq.
The area of triangle AOD is ¹/₂AO×DO sin(180-q)=¹/₂AO×DO sinq.
The area of triangle DOC is ¹/₂DO×CO sinq.
The area of triangle BOC is ¹/₂BO×CO sin(180-q)=¹/₂BO×CO sinq.
The area of the quadrilateral is the sum of these four triangles.

Area = 1 2 AO×BO sinq+ 1 2 AO×DO sinq+ 1 2 DO×CO sinq+ 1 2 BO×CO sinq = 1 2 [AO(BO+DO) + CO(DO+ BO)]sinq = 1 2 (AO×BD+ CO×BD)sinq = 1 2 AC×BD sinq

So we have proved that for a convex quadrilateral the area of the quadrilateral is given by half the product of the lengths of the diagonals multiplied by the sine of the angle between the diagonals.