How can I use nine-point circle to solve this concyclic problem I saw this problem on the Discord Math channel.

H is the orthocenter of △ABC. D, E and F are the foot of the altitudes of △ABC passing through A, B and C respectively.  Lines EF and BC intersect at R.  The line parallel to EF passing through D intersects AB and AC at P and Q respectively.  M is the midpoint of BC.  Prove that M, P, Q and R are concyclic.


Someone said that this can be proved using the nine-point circle, but I tried hard drawing figures to see things like ∠FEM = ∠BDF = ∠BAC.  I also know that BM = CM = EM = FM.  However, when P and M come together, I find the angle calculations hard because M is defined by a side length, and P is defined by an angle.  I can't see a way without using trigonometry functions.
I've derived on my own the following basic result that the orthic triangle △DEF admits H as its incenter.  I've used the result (1) in the previous paragraph.

I showed this by the converse of the intersecting chords theorem instead of nine-point circle.  I've chosen this approach for two reasons:

*

*I've got lost in the figures while calculating angles.

*M is defined by a side length, so it'll be easier to calculate side lengths.

     MD ⋅ DR = (0.5 BC − CD) (DC + CR)      ⋯⋯ ①
We calculate CR to eliminate R from this product.  I used Ceva's Theorem and Menelaus's Theorem to see that |BR : RC| = BD : DC.  This gives
RC : CB = RC : |BR − RC| = DC : (BD − DC) = CD : (BC − 2CD)
CR = BC ⋅ CD / (BC − 2CD)      ⋯⋯ ②
Substitute ② into ①:
     MD ⋅ DR
= (0.5 BC − CD) (DC + CR)
= 0.5 (BC − 2CD) [1 + BC / (BC − 2CD)] CD
= 0.5 [(BC − 2CD) + BC] CD
= (BC − CD) CD
= BD ⋅ CD      ⋯⋯ ③
Observe that B, Q, C and P are concyclic.




observation
reason




∠CQD = ∠AEF
corr. ∠s, EF // PQ


∠AEF = ∠ABC
result (1) in above screenshot


∴ ∠CQD = ∠DBP



i.e. B, Q, C and P are concyclic.
converse of ∠s in same seg.


BD ⋅ DC = PD ⋅ DQ
chord thm.


MD ⋅ DR = BD ⋅ DC
③


∴ MD ⋅ DR = PD ⋅ DQ



i.e. M, P, Q and R are concyclic.
converse of chord thm.




Q.E.D.
 A: Here is a way to proceed explicitly using similarities of triangles and the power of a point w.r.t. a circle. By symmetry, we may and do suppose that $\hat C$ is bigger than $\hat B$. Well, strictly bigger, else there is no $R$. The solution lives then in the following picture:

Here,
$
\widehat{BPQ} =
\widehat{BFE} = 180^\circ-\hat C=
\widehat{BCQ}
$, so $BPCQ$ cyclic, giving for the power of $D$ w.r.t. the circle $\odot(BPCQ)$
$$
\tag{$1$}
DP\cdot DQ=DB\cdot DC\ .
$$
(We want to compute $DM\cdot DR$ and show it is equal to the above. For this, $DR$ is re-expressed using similarities to "get closer" to lengths simpler related to $\Delta ABC$.)
Let $N\in AB$ be the point making $\Delta NBC$ isosceles. Then the two green angles in the pictures are congruent, same measure $\hat C-\hat B$, leading to $\Delta DRE\sim\Delta ACN$, so
$$
\tag{$2$}
\frac{DR}{AC}
=
\frac{DE}{AN}
=
\frac{\displaystyle\frac{DE}{AB}}{\displaystyle\frac{AN}{AB}}
=
\frac{\displaystyle\frac{DC}{AC}}{\displaystyle\frac{DM}{DB}}
=
\frac{DB\cdot DC}{AC\cdot DM}
\ .
$$
Simplifying with $AC$, we get
$$
\tag{$3$}
DM\cdot DR=DB\cdot DC\ .
$$
From $(1)$ and $(3)$ we get $DM\cdot DR=DP\cdot DQ$, so $MRPQ$ cyclic.
$\square$
A: A shorter solution:

*

*$B,P,C,Q$ are concyclic, since $APQ$ is similar to $AFE$, which is similar to $ACB$

*By the chords theorem $BD\cdot DC= PD\cdot DQ$, hence it is enough to show that $RD\cdot DM = BD\cdot DC$...

*... which is fairly simple by cross ratios, for instance.

