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 $\theta$ degrees, the angle between $DO$ and $CO$ is the
same. The angle between $AO$ and $DO$ is $180-\theta$ degrees, the
angle between $BO$ and $CO$ is the same.

The area of triangle $AOB$ is ${1\over 2}AO\times BO \sin
\theta$.

The area of triangle $AOD$ is ${1\over 2}AO\times DO \sin
(180-\theta)={1\over 2}AO\times DO \sin \theta$.

The area of triangle $DOC$ is ${1\over 2}DO\times CO \sin
\theta$.

The area of triangle $BOC$ is ${1\over 2}BO\times CO \sin
(180-\theta)={1\over 2}BO\times CO \sin \theta$.

The area of the quadrilateral is the sum of these four
triangles.

$$\eqalign{ {\rm Area}&={1\over 2}AO\times BO \sin
\theta+{1\over 2}AO\times DO \sin \theta+{1\over 2}DO\times CO \sin
\theta+{1\over 2}BO\times CO \sin \theta \cr &= {1\over
2}[AO(BO+DO) + CO(DO+ BO)]\sin \theta \cr &={1\over 2}(AO\times
BD+ CO\times BD)\sin \theta \cr &={1\over 2}AC\times BD \sin
\theta }$$

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.