No Right Angle Here

526 /www/nrich/html/content/97/09/six5/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Prove that the internal angle bisectors of a triangle will never be perpendicular to each other. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> 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 $AC$. <mdo:image align="absmiddle" alt="Diagram for No Right Angle Here" src="sept_triangle1.gif"></mdo:image> </mdoxml> 2 5 0 0 0 1 0 No Right Angle Here Prove that the internal angle bisectors of a triangle will never be perpendicular to each other. 2D Geometry, Shape and Space Angle properties of shapes Using, Applying and Reasoning about Mathematics Mathematical reasoning & proof 2D Geometry, Shape and Space Mixed triangles 2D Geometry, Shape and Space Bisection Secondary Mapping Document Geometrical reasoning US