### At a Glance

The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?

### A Sameness Surely

Triangle ABC has a right angle at C. ACRS and CBPQ are squares. ST and PU are perpendicular to AB produced. Show that ST + PU = AB

Points D, E and F are on the the sides of triangle ABC. Circumcircles are drawn to the triangles ADE, BEF and CFD respectively. What do you notice about these three circumcircles?

# No Right Angle Here

##### Stage: 4 Challenge Level:

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 $AC$.