Why is $DEFL$ a cyclic quadrilateral?

Question

Consider the following diagram. $A, B, C$ are vertices of a triangle and they each lie on a rectangular hyperbola. Let $H$ be the orthocenter of $\triangle ABC$ and let $L$ be the projection of $H$ onto the line $BC$. Let $D, E, F$ be the midpoints of the sides of the triangle, as shown below.

In the article, it was given ad verbatim

It is well known that the quadrilateral $DEFL$ is cyclic and its circumcircle is the Euler circle.

From the tone of this it seems that the cyclic nature of $DEFL$ is almost trivial, but I cannot see why this is the case. What am I missing here?

Answer 1

To show that $DEFL$ is cyclic, it is enough to show that $\angle DEF$ and $\angle DLF$ are supplementary. Note that $\angle DEF = \angle DBL$ (apply your favorite parallel lines or similar triangles argument), and that $\angle DLF$ and $\angle DLB$ are supplementary, so it suffices to show that $\angle DBL = \angle DLB$.

Since $L$ is the projection of the orthocenter onto $BC$, it is the foot of an altitude from $A$ onto $BC$, and hence $\angle ALB$ is right. This implies that $L$ lies on the circle where $AB$ is a diameter, i.e. the circle centered at the midpoint of $AB$ (i.e. $D$) containing the points $A$ and $B$. But that implies that $DL = DB$, so $\angle DLB = \angle DBL$, as desired.

Answer 2

Alternatively, you can prove it by showing that $\angle \, EFD = \angle \, ELD = \angle \, CAB = \alpha$ because if $\angle \, EFD = \angle \, ELD$ then $DEFL$ is cyclic.

The fact that $\angle\, EFD = \angle \, CAB = \alpha$ can be proved in many ways - triangle $EFD$ and $ABC$ are similar (in fact homothetic with respect to the centroid of $ABC$ and scaling coefficient $-1/2$) or $EFDA$ is a parallelogram, etc.

The fact that $\angle \, ELD = \angle \, CAB = \alpha$ can be seen by noticing that $LD$ is the median towards the hypotenuse $AB$ of the right-angled triangle $ABL$. Therefore $LD = AD$ and so triangle $ALD$ is isosceles and thus $\angle \, LAD = \angle \, ALD$. Analogously, the fact that $LE$ is the median towards the hypotenuse of the right-angled triangle $CAL$ implies that $\angle \, EAL = \angle \, ELA$. Adding the two identities together $\alpha = \angle \, CAB = \angle \, EAL + \angle LAD = \angle \, ELA + \angle \, DLA = \angle \, ELD$