Notes
four squares around a quadrilateral solution

Four Squares Around a Quadrilateral

Solution by Area of a Triangle

In the above diagram, the point $N$ is obtained by shifting $L$ parallel to $I J$ so that $N I$ has the same length as $L I$. This means that triangles $I L J$ and $I N J$ have the same area.

Angles $J \hat{I} N$ and $J \hat{I} L$ add up to $180^\circ$. This can be seen in various ways, one of which is that reflecting rectangle $L N P M$ through the line through $I$ takes triangle $N I P$ to $L I M$.

Angles $A \hat{I} B$ and $J \hat{I} L$ also add up to $180^\circ$ since the angles at a point add up to $360^\circ$. Since $A I$ and $N I$ have the same length, and $B I$ and $J I$ also have the same length, this means that triangles $A I B$ and $N I J$ are congruent and so have the same area.

Putting that together, triangles $A I B$ and $L I J$ have the same area. Following round the same argument for the other grey triangles, each grey triangle has the same area as a triangle formed by cutting the central quadrilateral along a diagonal. So the four grey regions have total area $20$.

Solution by the Sine Rule

This argument follows a similar line to the above, except that showing that triangles $L I J$ and $A I B$ have the same area is deduced from the sine rule. Since angles $L \hat{I} J$ and $A \hat{I} B$ add up to $180^\circ$, they have the same $sine$ and so triangles $A I B$ and $L I J$ have the same area.