Cyclic Quadrilaterals from Altitudes and Orthocenter

In $\triangle ABC$, altitudes are drawn from $A,B,$ and $C$ to $D,E,$ and $F$, respectively, as shown in the figure.

Cyclic quadrilaterals! There are several indications of the existence of these, but one of these is an important one we will use:

Opposite angles add up to $180$ degrees: If a quadrilateral is inscribed in a circle, the two arcs two opposite angles subtend sum to $360$ degrees, so the sum of the two angles is $\frac{360}{2}=180$ degrees.

The converse is also easily proved as if we assume a quadrilateral can be inscribed in a circle, and two opposite angles do not sum to $180$ degrees, the sum of the two arcs the angles subtend do not sum to $360$ degrees, thereby inciting a contradiction.

As we have noticed in this figure, there are right angles. Take for example quadrilateral $AEFH$. The two angles $\angle AFH$ and $\angle AEH$ are both right, so their sum is $180$ degrees. Thus, using the above statement, we have the quadrilateral is right.

Examining the same angles for angles $\angle B$ and $\angle C$, we see that quadrilaterals

$AEFH,$ $CDHE,$ $BFHD$

are cyclic. $\blacksquare$