This is the solution by Tim from Gravesend Grammar School.

Assume that two of the internal angle bisectors, $AM$ and $BN$, are perpendicular to each other, meeting at $X$, ie $\angle AXB = 90^\circ$.

In triangle $\Delta AXB$, $\angle AXB + \angle BAX + \angle XBA = 180^\circ$

so $\angle BAX + \angle XBA = 90^\circ$.

But $\angle XAC = \angle XAB$ and $\angle ABX = \angle XBC$

so the sum of the angles in the triangle is $2 ( \angle BAX + \angle XBA ) + \angle BCA = 180^\circ + \angle BCA$

so $\angle BCA = 0$, so $ABC$ is not a triangle as it only has two angles,

hence $AM$ and $BN$ are not perpendicular.

Note: What is happening here is that $BC$ is parallel to $AC$.

Angle properties of polygons. Pythagoras' theorem. Creating and manipulating expressions and formulae. Angles - points, lines and parallel lines. Circle properties and circle theorems. Visualising. Mathematical reasoning & proof. Regular polygons and circles. Similarity and congruence. Generalising.