Tom Clarke and James Frederick Well 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 θ degrees, the angle between DO and CO is the same. The angle between AO and DO is 180−θ degrees, the angle between BO and CO is the same.

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

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

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.