Show that the Points $E,H,Δ,Ζ$ belong to a circle

Question

Show that the Points $E,H,Δ,Ζ$ are homocyclic

Answer 1

$BA\Delta$ is a right triangle, therefore the vertices $B,\Delta,A$ lie on a circle centered at $Z$. Then, $ZB=Z\Delta$ (both are radii of the circle). Then the triangle $B\Delta Z$ is isosceles and angle $ZB\Delta$ = angle $B\Delta Z$. Then, angle $Z\Delta H$ is supplementary to angle $ZB\Delta$. On the other hand, since $Z, E, H$ are midpoints, $ZE$ is parallel to $B\Gamma$ and $EH$ is parallel to $AB$. In other words, $BHEZ$ is a parallelogram and hence angle $ZBH$ = angle $ZEH$. Conclusion: the opposite angles of the quadrilateral $Z\Delta HE$ are supplementary and its vertices lie on a circle.

Answer 2

Since Z, E, and H are the side midpoints, ZE || BC and HE = BZ. Since ABΔ is a right triangle with Z the center of the circle circumscribing ABΔ, ZΔ = BZ = HE.

Because ZE || HΔ and ZΔ = HE, ZBHΔ is a symmetric trapezoid, with supplementary opposite angles. Therefore, ZBHΔ is cyclic.

Answer 3

With different notation ...

$$\left.\begin{align} \square AEDF \;\text{is a}\;\parallel\text{-ogram} &\implies \angle D \cong \angle A \\ \overline{EF}\;\text{is}\;\perp\text{bisector of}\;\; \overline{AA'} &\implies \angle A'\cong\angle A \end{align}\right\} \implies \angle D \cong \angle A' \stackrel{\star}{\implies} A'DEF \;\text{concyclic} $$ $\star$: An aspect of the Inscribed Angle Theorem. $\square$

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Note. The resulting circle obviously also passes through the feet of the altitudes from $B$ and $C$. Perhaps less-obviously (though readily verified), it passes through the midpoints of segments joining the orthocenter to $A$, $B$, $C$. This is the well-known Nine-Point Circle.