User:Bawolff/math steps

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Let ${\rm {AB\Gamma }}$ be a triangle. The altitudes ${\rm {BE}}$ and ${\rm {\Gamma Z}}$ intersect at ${\rm {H}}$. We will show that also ${\rm {AH}}$ is perpendicular to ${\rm {B\Gamma }}$, i.e. an altitude of ${\rm {AB\Gamma }}$.

• The quadrilateral ${\rm {BZE\Gamma }}$ is cyclic, because $\angle {\rm {BZ\Gamma }}=\angle {\rm {\Gamma EB}}=90^{\circ }$. So, also ${\widehat {\rm {ZEB}}}={\widehat {\rm {Z\Gamma B}}}={\hat {\omega }}$ (subtended by the same arc).
• The quadrilateral ${\rm {AZHE}}$ is cyclic because opposite angles ${\widehat {\rm {AEH}}}$ and ${\widehat {\rm {AZH}}}$ are supplementary (${\widehat {\rm {AEH}}}+{\widehat {\rm {AZH}}}=90^{\circ }+90^{\circ }=180^{\circ }$). So, angles ${\widehat {\rm {ZAH}}}$ and ${\widehat {\rm {ZEH}}}={\hat {\omega }}$ are equal.
• The angles ${\widehat {\rm {AHZ}}}$ and ${\widehat {\rm {\Delta H\Gamma }}}$ are equal as vertical angles, i.e. ${\widehat {\rm {AHZ}}}={\widehat {\rm {\Delta H\Gamma }}}={\hat {\varphi }}$. In the right triangle ${\rm {AHZ}}$, we have that ${\hat {\varphi }}+{\hat {\omega }}=90^{\circ }$, therefore in the triangle ${\rm {H\Delta \Gamma }}$ it holds that ${\widehat {\rm {A\Delta \Gamma }}}=90^{\circ }$.

We conclude that, ...

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This is based on a script by User:Dimitri131