Problem #PRU-5086

Problem

Consider the following "proof" that any triangle is equilateral: Given a triangle $ABC$, we first prove that $AB=AC$. First let’s draw the bisector of the angle $\mathrm{\angle }A$. Now draw the perpendicular bisector of segment $BC$, denote by $D$ the middle of $BC$ and by $O$ the intersection of these lines. See the diagram

Draw the lines $OR$ perpendicular to $AB$ and $OQ$ perpendicular to $AC$. Draw lines $OB$ and $OC$. Then the triangles, $RAO$ and $QAO$ are equal, since we have equal angles $\mathrm{\angle }ORA=\mathrm{\angle }OQA=90°,$ and $\mathrm{\angle }RAO=\mathrm{\angle }QAO,$ and the common side $AO$. On the other hand the triangles $ROB$ and $QOC$ are also equal since the angles $\mathrm{\angle }BRO=\mathrm{\angle }CQO=90°$, the hypotenuses $BO=OC$ the legs $RO=OQ$. Thus, $AR=AQ,$ $RB=QC,$ and $AB=AR+RB=AQ+QC=AC.$ Q.E.D.

As a corollary, one can show that all the triangles are equilateral, by showing that $AB=BC$ in the same way.