Question 68
Let $\triangle ABC$ be isosceles, with $AB = AC$, and $P$, $Q$ be points on the side $AC$ so that $m(\widehat{ABP}) = m(\widehat{PBQ}) = m(\widehat{QBC})$. If $[AD]$ is an altitude, $D \in BC$, $BP \cap AD = \{M\}$, $BQ \cap AD = \{N\}$ and $\triangle ABN$ is isosceles, prove that:
a) $M$ is the orthocenter of triangle $ABC$;
b) $MN = \frac{AB}{AD}(AB - AD)$.
a) $M$ is the orthocenter of triangle $ABC$;
b) $MN = \frac{AB}{AD}(AB - AD)$.