Simson Line
Before starting let us go through some notaions for simplicity.
$ (ABC) := $ circumcircle of $ \triangle ABC $
$x \cap y := $ intersection of object $x$ and $y$
To avoid configuration issues we will use the notion of directed angles in most of the cases.
$ \measuredangle ABC := \angle ABC \ ( \mbox{ mod } 180^{\circ} \ ) $
To learn more about directed angles one can concern this.
Observe that quadrilaterals $ AYPZ, BZPX , CXPY $ are all cyclic quadrilaterals.
Therefore $ \measuredangle PYZ = \measuredangle PAZ = \measuredangle PAB = \measuredangle PCB = \measuredangle PCX = \measuredangle PYX $
Hence $ X,Y$ and $Z$ are collinear, which completes our lemma.
But the Simson line configuration does not stops here, there are many more interesting properties!
Lemma $2$ (Simson Line Bisection) : Let $ ABC $ be a triangle with orthocenter $H$. If $P$ is a point on $(ABC)$ then its Simson line bisects $ \overline{ \rm PH } $
Let us extend our configuration a little bit more.
Denote by $H$ the orthocenter of triangle $ABC$ and let $PX$ meet $(ABC)$ again at a point $K$, and let line $AH$ intersect the Simson line at the point $L$. Let $D$ be the foot of perpendicular from $A$ to $BC$ and let $K'$ be reflection of $K$ about $BC$.
Claim 1 : The Simson line is parallel to $ AK $.
Proof :
We have $ \measuredangle PXY = \measuredangle PCY = \measuredangle PCA = \measuredangle PKA $
$ \therefore XY \parallel AK $, which completes our proof.
$ \boxed{ \therefore LAKX \mbox{ is a parallelogram.} }$
Claim 2 : $K'$ is orthocenter of $ \triangle PBC $.
Proof :
Let $ BK' $ intersect $ CP $ at $E$.
We have $ BK' = BK $
$\Rightarrow \measuredangle XPC = \measuredangle KPC = \measuredangle KBC = \measuredangle KBX = \measuredangle XBK' = \measuredangle CBE $
$ \therefore \measuredangle BEC = \measuredangle BCE - \measuredangle CBE = \measuredangle BCP - \measuredangle XPC = \measuredangle XCP - \measuredangle XPC = 90^{\circ} $
$ \therefore BK' \perp CP \Rightarrow K' $ is orthocenter of $ \triangle PBC $.
Claim 3 : $ LHXP $ is a parallelogram.
Proof :
First we will show that $ AHK'P $ is parallelogram.
Let $ AD \cap (ABC) = H' $
By Claim 2, we get $ H' $ is reflection of $H$ about $BC$
$ \therefore H'K $ is reflection of $HK'$ about $BC$, thus $ H'K = HK' $
Now $ AH' \parallel PK $ and $ AH'KP $ is cyclic $ \Rightarrow AH'KP $ is an isosceles trapezoid.
$ \therefore AP = H'K = HK' \Rightarrow AHK'P $ is a parallelogram.
Now $ LA = XK $ ( as $ LAKX$ is a parallelogram ) and $ AH = PK' $ ( as $ AHK'P $ is a parallelogram ).
From here we deduce that $ LH = PX $.
Thus, as $ LH \parallel PX $, we get $ LHXP $ is a parallelogram.
Thus from here we deduce that $ XL $ bisects $ PH $ ( as $LHXP $ is a parallelogram )
Hence the Simson line bisects $PH$, which completes our beautiful lemma.
Well let us dwell into more interesting properties of Simson lines.
Lemma $3$ : Let $ ABC $ be a triangle. Let $X$ be midpoint of arc $BC $ containing $A$. Similarly define $ Y$ and $ Z$. Then the Simson line of $X,Y$ and $Z$ with respect to $ \triangle ABC $ are concurrent.
Let $ D,E $ and $ F$ be midpoints of $ BC, CA $ and $ AB $ respectively. Let $ XD \cap (ABC) = K $ and $ XM \perp CA $ and $ XN \perp AB $
Observe that Simson line of $X$ with respect to $ \triangle ABC $ is $ DM $
Claim : $DM$ is internal angle bisector of $ \angle FDE $.
Proof :
By Lemma $2$ , Claim $1$ we have $ AK \parallel DM $
By Lemma $2$ , Claim $1$ we have $ AK \parallel DM $
$ \therefore \angle DME = \angle KAC = \frac{1}{2} \angle A $
And we know that $ \angle MED = \angle MEF + \angle FED = \angle C + \angle B $
$ \therefore \angle EDM = 180^{\circ} - \angle DME - \angle MED = \frac{1}{2} \angle A = \frac{1}{2} \angle FDE $, which completes our claim.
Now as the angle bisectors of a triangle are concurrent, we deduce that Simson line of $X,Y $ and $Z$ with respect to $ \triangle ABC $ are concurrent.
Lemma $4$ : Let $ABCD$ be a cyclic quadrilateral. Then the Simson lines of $A,B,C$ and $D$ with respect to triangles $ BCD, CDA, DAB $ and $ABC$ respectively are concurrent.
Proof :
The proof of this lemma is quite easy if one knows Lemma $2$.
Let $H_a ,H_b, H_c $ and $ H_d$ be the orthocenters of triangles $ BCD, CDA , DAB $ and $ABC$ respectively.
Now we have already proved that if $ ABCD $ is any cyclic quadrilateral and $ H_a $ and $ H_b $ are orthocenters of triangle $ BCD $ and $ CDA $ respectively, then $ ABH_aH_b $ is a parallelogram ( the proof was done in Lemma $2$ , Claim $3$ )
Thus $ AH_a \cap BH_b $ is midpoint of both of them, and again by Lemma $2$, Simson lines of $A$ and $B$ with respect to triangles $ BCD$ and $CDA$ bisects $ BH_b $ and $ AH_a $, hence the Simson line of $A$ and $B$ with respect to triangles $BCD$ and $CDA$ passes through midpoint of $BH_b$.
Similarly looking at parallelograms $ BCH_bH_c $ and $ BDH_bH_d $ we get Simson lines of $C$ and $D$ with respect to triangles $ DAB $ and $ ABC $ respectively passes through midpoint of $BH_b $.
Hence the Simson lines of $A,B,C$ and $D$ with respect to triangles $ BCD, CDA, DAB $ and $ABC$ respectively are concurrent, which completes our lemma.
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