#### Hide answer & extension

Draw three more squares and add these lines (I have coloured the angles to make equal angles clearer):

Triangles \(ACE\), \(LDK\) and \(IKE\) are congruent, so angle \(KDL\) is equal to \(\beta\).

The congruence of these triangles tells us that angles \(DKL\) and \(EKI\) add up to a right angle, so angle \(EKD\) is also a right angle.

The congruence of the triangles also tells us that \(KD\) and \(KE\) are the same length and so angle \(EDK\) is the angle in an isosceles right-angled triangle. \(\alpha\) is also the angle in an isosceles right-angled triangle, so these two angles are equal.

Therefore \(\alpha+\beta+\gamma=90^\circ\).

#### Extension

The diagram shows three rhombuses with diagonals drawn on and three angles labelled.

What is the value of \(\alpha+\beta+\gamma\)?