parmenides51 wrote: In the exterior of the triangle $ABC$ with $m(\angle B) > 45^o$, $m(\angle Q) >45°^o$ one constructs the right isosceles triangles $ACM$ and $ABN$ such that $m(\angle CAM) = m(\angle BAN) = 90^o$ and, in the interior of $ABC$, the right isosceles triangle $BCP$, with $m(\angle P) = 90^o$. Show that the triangle $MNP$ is a right isosceles triangle. $m(\angle C) >45°^o$ Given $\triangle ABC\ :\ A(0,a),B(-b,0),C(c,0)$. Points $M(a,a+c),N(-a,a+b),P(\frac{c-b}{2},\frac{c+b}{2})$

vanstraelen wrote: ... $m(\angle C) >45°^o$ ... indeed, corrected , tnx