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