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Step by step proofs (using calculator)

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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. that it is 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 $\angle {\rm {ZEB}}=\angle {\rm {Z\Gamma B}}={\hat {\omega }}$ (as angles subtended by the same arc).
• The quadrilateral ${\rm {AZHE}}$ is cyclic because opposite angles $\angle {\rm {AEH}}$ and $\angle {\rm {AZH}}$ are supplementary (since $\angle {\rm {AEH}}+\angle {\rm {AZH}}=90^{\circ }+90^{\circ }=180^{\circ }$). So, angles $\angle {\rm {ZAH}}$ and $\angle {\rm {ZEH}}={\hat {\omega }}$ are equal.
• The angles $\angle {\rm {AHZ}}$ and $\angle {\rm {\Delta H\Gamma }}$ are equal as vertical angles, i.e. $\angle {\rm {AHZ}}=\angle {\rm {\Delta H\Gamma }}={\hat {\varphi }}$. 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 $\angle {\rm {A\Delta \Gamma }}=90^{\circ }$.
We conclude that the three altitudes pass through the same point.