In this post, I will explore many results regarding the altitudes, the medians, the orthocenter and the centroid of a triangle $\triangle ABC$ .

I will denote by $H_A, H_B, H_C$ the feet of the altitudes with respect to $A, B, C$ and $M_A, M_B, M_C$ the midpoints of sides $a, b, c$ . Also let $H$ and $G$ denote the orthocenter and centroid of $\triangle ABC$ .

Let's start with a well-known result:

1:Nine-point circle: The points $H_A, H_B, H_C, M_A, M_B, M_C$ are all concyclic.

Proof: Since $\triangle ABH_A$ is right and $M_C$ is the midpoint of $AB$ , we have that $\triangle M_CBH_A$ is isosceles, thus $\angle BH_AM_C = \angle B and \angle M_CH_AM_A = 180 - \angle B = 180 - \angle M_CM_BM_A$ . Therefore, $H_A, M_A, M_B, M_C$ are concyclic, and similarly all six points lie on a circle.

Remark. The midpoints of $AH, BH, CH$ are also in this circle, thus the name nine-point circle. The proof is analogous angle-chasing.

2:Prove that the ray $H_AG$ intersects the circumcircle of $\triangle ABC$ at a point $T'_A$ such that $AT'_A$ is parallel to $BC$ .

$[asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.55, xmax = 13.48, ymin = -3.63, ymax = 4.32; /* image dimensions */ pen ccwwqq = rgb(0.8,0.4,0); pen ttttff = rgb(0.2,0.2,1); pen ttzzqq = rgb(0.2,0.6,0); pen ffcctt = rgb(1,0.8,0.2); pen cccccc = rgb(0.8,0.8,0.8); draw((2.24,3.52)--(0.66,-1.34)--(7.36,-1.36)--cycle, ccwwqq); /* draw figures */ draw((2.24,3.52)--(0.66,-1.34), ccwwqq); draw((0.66,-1.34)--(7.36,-1.36), ccwwqq); draw((7.36,-1.36)--(2.24,3.52), ccwwqq); draw(circle((4.01,0.26), 3.72), ttttff); draw(circle((3.12,0.28), 1.86), ttttff); draw((4.8,1.08)--(4.01,-1.35), ttzzqq); draw((4.01,-1.35)--(1.45,1.09), ttzzqq); draw((1.45,1.09)--(4.8,1.08), ttzzqq); draw((2.24,3.52)--(5.81,3.51), red); draw((2.24,3.52)--(2.23,-1.34), ffcctt); draw((4.01,-1.35)--(2.24,3.52), cccccc); draw((1.45,1.09)--(7.36,-1.36), cccccc); draw((4.8,1.08)--(0.66,-1.34), cccccc); draw((5.81,3.51)--(1.43,-2.42), linetype("4 4") + red); draw((4.02,1.08)--(4.01,-1.35)); /* dots and labels */ dot((2.24,3.52),dotstyle); label("$A$", (2.28,3.59), NE * labelscalefactor); dot((0.66,-1.34),dotstyle); label("$B$", (0.55,-1.54), NW * 2); dot((7.36,-1.36),dotstyle); label("$C$", (7.43,-1.58), NE * labelscalefactor); dot((4.8,1.08),dotstyle); label("$M_B$", (4.97,0.96), NE * labelscalefactor); dot((1.45,1.09),dotstyle); label("$M_C$", (1.18,1.01), W *0.5); dot((4.01,-1.35),dotstyle); label("$M_A$", (4,-1.53), S); dot((3.42,0.27),dotstyle); label("$G$", (3.23,0.36), NE * labelscalefactor); dot((2.23,-1.34),dotstyle); label("$H_A$", (2.26,-1.55), S); dot((1.43,-2.42),dotstyle); label("$T_A$", (1.34,-2.19), NE * labelscalefactor); dot((5.81,3.51),dotstyle); label("$T'_A$", (5.78,3.29), NE*2); dot((4.01,0.26),dotstyle); label("$O$", (4.06,0.33), NE * labelscalefactor); dot((4.02,1.08),dotstyle); label("$H'_A$", (3.76,1.2), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$

Proof: Let $\mathcal{H}$ denote the homothety centered at $G$ which sends $\triangle ABC$ to $\triangle M_AM_BM_C$ . Then the image of $T'_A$ is the intersection of the nine-point circle of $\triangle ABC$ and $GH_A$ , which is just $H_A$ . But then $H_A M_A \parallel AT'_A$ , so $BC \parallel AT'_A$ .

3:Let $\mathcal{C}$ denote the circle passing through $M_B$ , $M_C$ tangent to the circumcircle $\omega$ of $ABC$ at a point $T_A \neq A$ . Then $T_A, H_A, G$ are collinear.

$[asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.59, xmax = 13.85, ymin = -4.12, ymax = 4.54; /* image dimensions */ pen ccwwqq = rgb(0.8,0.4,0); pen ttttff = rgb(0.2,0.2,1); pen ttzzqq = rgb(0.2,0.6,0); pen ffcctt = rgb(1,0.8,0.2); pen cccccc = rgb(0.8,0.8,0.8); draw((2.14,3.44)--(0.66,-1.34)--(7.36,-1.36)--cycle, ccwwqq); /* draw figures */ draw((2.14,3.44)--(0.66,-1.34), ccwwqq); draw((0.66,-1.34)--(7.36,-1.36), ccwwqq); draw((7.36,-1.36)--(2.14,3.44), ccwwqq); draw(circle((4.01,0.24), 3.71), ttttff); draw(circle((3.07,0.25), 1.85), ttttff); draw((4.75,1.04)--(4.01,-1.35), ttzzqq); draw((4.01,-1.35)--(1.4,1.05), ttzzqq); draw((2.14,3.44)--(5.91,3.43), red); draw((2.14,3.44)--(2.12,-1.34), ffcctt); draw((4.01,-1.35)--(2.14,3.44), cccccc); draw((1.4,1.05)--(7.36,-1.36), cccccc); draw((4.75,1.04)--(0.66,-1.34), cccccc); draw((5.91,3.43)--(1.34,-2.33), linetype("4 4") + red); draw((4.02,1.04)--(4.01,-1.35)); draw(circle((3.07,-0.67), 2.4), cccccc); draw(circle((3.08,1.84), 1.85), cccccc); draw((2.14,3.44)--(-1.92,1.06)); draw((-1.92,1.06)--(1.34,-2.33)); draw((-1.92,1.06)--(7.64,1.03), ttzzqq); draw((5.91,3.43)--(0.4,1.05), red); draw((5.91,3.43)--(7.64,1.03), red); /* dots and labels */ dot((2.14,3.44),dotstyle); label("$A$", (2.18,3.51), NE * labelscalefactor); dot((0.66,-1.34),dotstyle); label("$B$", (0.54,-1.56), W); dot((7.36,-1.36),dotstyle); label("$C$", (7.43,-1.59), NE * labelscalefactor); dot((4.75,1.04),dotstyle); label("$M_B$", (4.93,0.91), N*1.5); dot((1.4,1.05),dotstyle); label("$M_C$", (1.11,0.96), N); dot((4.01,-1.35),dotstyle); label("$M_A$", (4,-1.54), S); dot((3.39,0.25),dotstyle); label("$G$", (3.18,0.34), NE * labelscalefactor); dot((2.12,-1.34),dotstyle); label("$H_A$", (2.16,-1.57), S*0.25); dot((1.34,-2.33),dotstyle); label("$T_A$", (1.23,-2.07), S*4); dot((5.91,3.43),dotstyle); label("$T'_A$", (5.88,3.19), NE*2); dot((4.01,0.24),dotstyle); label("$O$", (4.06,0.32), NE * labelscalefactor); dot((4.02,1.04),dotstyle); label("$H'_A$", (3.74,1.16), N); dot((-1.92,1.06),dotstyle); label("$P_A$", (-1.98,1.22), NE * labelscalefactor); dot((7.64,1.03),dotstyle); label("$T_2$", (7.71,1.18), NE * labelscalefactor); dot((0.4,1.05),dotstyle); label("$T_1$", (0.17,1.19), W); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$

Proof: Let $\mathcal{C}_2$ denote the circumcircle of $AM_CM_B$ , which is also tangent to $\omega$ . Let $P_A$ be the radical center of $\mathcal{C}, \mathcal{C}_2, \omega$ , and let $T_1, T_2$ be the intersections of $M_BM_C$ and $\omega$ , as above. Then since $P_AA$ and $P_AT_A$ are tangent to $\omega, AT_1T_AT_2$ is a harmonic quadrilateral. Let $T_A^*$ denote the second intersection of $T'_AG$ and $\omega$ , and $H'_A$ denote the image of $H_A$ under the homothety $\mathcal{H}$ which sends $\triangle ABC$ to $\triangle M_AM_BM_C$ . Then $M_AH'_A \perp T_1T_2$ , so $OH'_A \perp T_1T_2$ and $H'_A$ is the midpoint of $T_1T_2$ . We have shown in (2) that $AT'_A \parallel T_1T_2$ , so $T'_A(T_2, T_1; H'_A, A)$ is a harmonic pencil, which implies that $T'_A(T_2, T_1; T_A^*, A)$ is a harmonic pencil, and $AT_1T_A^*T_2$ is a harmonic quadrilateral $\implies T_A^* \equiv T_A$ , as desired.

Now, a simple but very useful lemma about the orthocenter:

4: Let $A'$ be the point diametrically opposite to $A$ w.r.t. $\omega$ . Prove that the ray $HA'$ intersects $BC$ at $M_A$ , and that $M_A$ is also the midpoint of $HA'$ .

$[asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 24.5, ymin = -8, ymax = 6.3; /* image dimensions */ pen ccwwqq = rgb(0.8,0.4,0); pen ffcctt = rgb(1,0.8,0.2); draw((4,4.46)--(2,-3.04)--(10.76,-3.2)--cycle, ccwwqq); /* draw figures */ draw((4,4.46)--(2,-3.04), ccwwqq); draw((2,-3.04)--(10.76,-3.2), ccwwqq); draw((10.76,-3.2)--(4,4.46), ccwwqq); draw(circle((6.43,-0.21), 5.26)); draw((2.54,-1.01)--(10.76,-3.2), ffcctt); draw((6.85,1.24)--(2,-3.04), ffcctt); draw((2,-3.04)--(8.87,-4.87)); draw((10.76,-3.2)--(8.87,-4.87)); draw((3.89,-1.37)--(8.87,-4.87)); draw((4,4.46)--(8.87,-4.87), red); draw((3.86,-3.07)--(4,4.46), ffcctt); /* dots and labels */ dot((4,4.46),dotstyle); label("$A$", (3.82,4.58), NE * labelscalefactor); dot((2,-3.04),dotstyle); label("$B$", (1.7,-3.28), W*0.5); dot((10.76,-3.2),dotstyle); label("$C$", (10.94,-3.34), NE * labelscalefactor); dot((2.54,-1.01),dotstyle); label("$H_C$", (2.16,-0.88), W*0.5); dot((6.85,1.24),dotstyle); label("$H_B$", (6.92,1.36), NE * labelscalefactor); dot((3.86,-3.07),dotstyle); label("$H_A$", (3.98,-2.92), NE * labelscalefactor); dot((6.38,-3.12),dotstyle); label("$M_A$", (6.54,-2.98), NE * labelscalefactor); dot((3.89,-1.37),dotstyle); label("$H$", (3.98,-1.24), NE * labelscalefactor); dot((8.87,-4.87),dotstyle); label("$A'$", (9,-5.12), NE * labelscalefactor); dot((6.43,-0.21),dotstyle); label("$O$", (6.52,-0.08), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$

Proof: I claim that $BHCA'$ is a parallelogram. Indeed, $BH \parallel A'C$ since it is immediate that $\angle HBC = \angle A'CB = 90 - \angle C$ . Analogously, $HC \parallel BA'$ since we have $\angle BCH = \angle CBA' = 90- \angle B$ . Therefore, $BHCA'$ is a parallelogram, thus $HA'$ passes through $M_A$ , and $M_A$ is the midpoint of $HA'$ .

5: (USA TSTST 2011 #4 restatement) Let rays $M_B H$ and $M_C H$ intersect the circumcircle $\omega$ of $\triangle ABC$ at $B_1$ and $C_1$ respectively. Then $P_A, B_1, C_1$ are collinear, where $P_A$ is defined as in (3).

$[asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.2, xmax = 15.2, ymin = -4.83, ymax = 5.3; /* image dimensions */ pen ccwwqq = rgb(0.8,0.4,0); pen ttttff = rgb(0.2,0.2,1); pen ttzzqq = rgb(0.2,0.6,0); pen cccccc = rgb(0.8,0.8,0.8); draw((2.04,3.81)--(0.66,-1.34)--(7.36,-1.36)--cycle, ccwwqq); /* draw figures */ draw((2.04,3.81)--(0.66,-1.34), ccwwqq); draw((0.66,-1.34)--(7.36,-1.36), ccwwqq); draw((7.36,-1.36)--(2.04,3.81), ccwwqq); draw(circle((4.02,0.52), 3.84), ttttff); draw(circle((3.02,0.3), 1.92), ttttff); draw((4.7,1.23)--(4.01,-1.35), ttzzqq); draw((4.01,-1.35)--(1.35,1.24), ttzzqq); draw(circle((3.03,2.17), 1.92), cccccc); draw((2.04,3.81)--(-2.23,1.25)); draw((-2.23,1.25)--(7.79,1.22), ttzzqq); draw(circle((3.02,-0.84), 2.66)); draw((4.7,1.23)--(7.37,2.38), linetype("4 4")); draw((1.35,1.24)--(0.67,2.4), linetype("4 4")); draw((0.36,-0.65)--(4,-3.32)); draw((0.36,-0.65)--(-2.23,1.25), linetype("4 4") + red); draw((0.67,2.4)--(7.37,2.38), ttzzqq); draw((0.36,-0.65)--(4.7,1.23)); draw((1.35,1.24)--(4,-3.32)); /* dots and labels */ dot((2.04,3.81),dotstyle); label("$A$", (2.1,3.89), N * labelscalefactor); dot((0.66,-1.34),dotstyle); label("$B$", (0.53,-1.59), W * labelscalefactor); dot((7.36,-1.36),dotstyle); label("$C$", (7.44,-1.63), NE * labelscalefactor); dot((4.7,1.23),dotstyle); label("$M_B$", (4.92,1.07), E *0.2); dot((1.35,1.24),dotstyle); label("$M_C$", (0.88,1.09), N); dot((4.01,-1.35),dotstyle); label("$M_A$", (4,-1.57), S * labelscalefactor); dot((-2.23,1.25),dotstyle); label("$P_A$", (-2.31,1.44), NE * labelscalefactor); dot((7.79,1.22),dotstyle); label("$T_2$", (7.88,1.39), NE * labelscalefactor); dot((0.25,1.24),dotstyle); label("$T_1$", (-0.01,1.4), NE * labelscalefactor); dot((2.03,0.07),dotstyle); label("$H$", (2.01,0.18), NE * labelscalefactor); dot((0.36,-0.65),dotstyle); label("$B_1$", (0.1,-0.92)); dot((4,-3.32),dotstyle); label("$C_1$", (3.9,-3.54), S * labelscalefactor); dot((0.67,2.4),dotstyle); label("$C'$", (0.51,2.52), N * labelscalefactor); dot((7.37,2.38),dotstyle); label("$B'$", (7.42,2.46), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$

Proof: Let $B'$ and $C'$ denote the antipodes of $B$ and $C$ w.r.t. $\omega$ . From (4) we have that $M_B$ and $M_C$ are the midpoints of $HB'$ and $HC'$ , thus $B'C' \parallel M_BM_C$ , so $\angle C_1M_CM_B = \angle C_1C'B' = \angle C_1B_1M_B$ , hence $M_BM_CB_1C_1$ is cyclic with circumcircle $\mathcal{C}_1$ . Let $\mathcal{C}_2$ be the circumcircle of $AM_BM_C$ . Then $P_A$ is the radical center of $\omega, \mathcal{C}_1, \mathcal{C}_2$ , therefore $P_A, B_1, C_1$ are collinear, as desired.

6: (APMO 2012 #4 restatement) Let $\triangle ABC$ be an acute triangle with circumcircle $\omega$ . Denote by $A_1$ the intersection of the ray $M_AH$ with $\omega$ and by $A_2$ the second intersection of $A_1H_A$ and $\omega$ . Prove that the quadrilateral $ABA_2C$ is harmonic.

First, I will prove a lemma about harmonic quadrilaterals:

$\bullet$ $\blacktriangleright$ Let $ABCD$ be a harmonic quadrilateral with circumcircle $\omega$ and let $P$ be any point in the plane different from $A,B,C,D$ . Denote by $A', B', C', D'$ the second intersections of $PA, PB, PC, PD$ and $\omega$ . Then $A'B'C'D'$ is also a harmonic quadrilateral.

Proof of lemma: Using similar triangles, it's not hard to find that:
$A'B'\cdot C'D' = A'D'\cdot B'C' = \frac{AB \cdot CD \cdot \mathcal{P}(P)^2}{EA \cdot EB \cdot EC \cdot ED}$ ,
where $\mathcal{P}(P)$ denotes the power of $P$ w.r.t. $\omega$ . Thus $A'B'C'D'$ is harmonic.

Main Proof: Let $A'$ denote the point diametrically opposite to $A$ w.r.t. $\omega$ , and $H'$ the reflection of $H$ w.r.t. $BC$ . It is easy to prove that $H' \in \omega$ , and since $M_A$ is the midpoint of $HA'$ , we have $A'H' \parallel BC \implies A'(B,C;M_A,H')$ is harmonic, so $CH'BA_1$ is a harmonic quadrilateral. Applying the above lemma to $CH'BA_1$ and taking $P \equiv H_A$ gives the desired result.

7: (OBM (Brazil MO) 2011 #5 restatement) Let $\triangle ABC$ be an acute triangle with circumcircle $\omega$ . Let the circumcircle of $AH_BH_C$ intersect $\omega$ at $F$ . Prove that the angle bisectors of $\angle BFC$ and $\angle BHC$ intersect at a point on side $BC$ .

$[asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 24.5, ymin = -8, ymax = 6.3; /* image dimensions */ pen ccwwqq = rgb(0.8,0.4,0); pen zzzzzz = rgb(0.6,0.6,0.6); pen ffcctt = rgb(1,0.8,0.2); draw((5.54,4.3)--(4.06,-2.76)--(10.8,-2.78)--cycle, ccwwqq); /* draw figures */ draw(circle((7.44,0.22), 4.5), blue); draw((5.54,4.3)--(4.06,-2.76), ccwwqq); draw((4.06,-2.76)--(10.8,-2.78), ccwwqq); draw((10.8,-2.78)--(5.54,4.3), ccwwqq); draw(circle((5.53,1.31), 2.99), zzzzzz); draw(circle((6.49,0.77), 3.66), zzzzzz); draw((10.8,-2.78)--(5.52,-3.86)); draw((5.52,-3.86)--(4.06,-2.76)); draw((4.06,-2.76)--(2.95,-0.2)); draw((8.39,0.46)--(4.06,-2.76), ffcctt); draw((5.54,4.3)--(5.52,-3.86), ffcctt); draw((9.34,-3.87)--(2.95,-0.2)); draw((4.34,-1.43)--(10.8,-2.78), ffcctt); draw((5.54,4.3)--(2.95,-0.2)); /* dots and labels */ dot((5.54,4.3),dotstyle); label("$A$", (5.62,4.42), NE * labelscalefactor); dot((4.06,-2.76),dotstyle); label("$B$", (4.14,-2.64), NE * labelscalefactor); dot((10.8,-2.78),dotstyle); label("$C$", (10.88,-2.66), NE * labelscalefactor); dot((4.34,-1.43),dotstyle); label("$H_C$", (4.42,-1.3), NE * labelscalefactor); dot((8.39,0.46),dotstyle); label("$H_B$", (8.48,0.58), NE * labelscalefactor); dot((2.95,-0.2),dotstyle); label("$F \equiv A_1$", (2.56,-0.18), S*3); dot((5.52,-1.67),dotstyle); label("$H$", (5.6,-1.56), NE * labelscalefactor); dot((5.52,-3.86),dotstyle); label("$H'$", (5.6,-3.74), NE * labelscalefactor); dot((7.43,-2.77),dotstyle); label("$M_A$", (7.52,-2.66), NE * labelscalefactor); dot((5.52,-2.76),dotstyle); label("$H_A$", (5.6,-2.64), NE * labelscalefactor); dot((9.34,-3.87),dotstyle); label("$A'$", (9.42,-3.74), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$

Proof: From the angle bisector theorem, the problem statement is equivalent to proving that $\frac{BH}{CH} = \frac{BF}{CF}$ . This looks a lot like a harmonic quadrilateral, but of course $H \notin \omega$ . Fortunately, we can take the reflection $H'$ of $H$ w.r.t. $BC$ , so that $\frac{BH'}{CH'} = \frac{BH}{CH}$ and $H' \in \omega$ . Now, it remains to prove that $CH'BF$ is a harmonic quadrilateral. But in (6) we have shown that $CH'BA_1$ is a harmonic quadrilateral, where $A_1$ is the intersection of the ray $M_AH$ and $\omega$ .
Thus, it suffices to prove $F \equiv A_1 \iff A_1, H_C, H_B, A$ are concyclic. Notice that $H$ is clearly in the circumcircle of $AH_BH_C$ . But then $\angle AA_1H = \angle AH_BH = 90$ , hence $A_1$ is also in the circumcircle of $AH_BH_C$ , so $F \equiv A_1$ , as desired.

8: Let $A, B, C,D$ be four concyclic points. Let $s_a$ denote the Simson line of $A$ w.r.t. $BCD$ , and define $s_b, s_c, s_d$ similarly. Let $h_a$ denote the line connecting $A$ to the orthocenter of $BCD$ , and define $h_b, h_c, h_d$ similarly. Prove that the eight lines $h_a, h_b, h_c, h_d, s_a, s_b, s_c, s_d$ are all concurrent.

$[asy] import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.44, xmax = 11.65, ymin = -1.79, ymax = 5.71; /* image dimensions */ pen ccwwqq = rgb(0.8,0.4,0); pen wwwwww = rgb(0.4,0.4,0.4); pen ffcctt = rgb(1,0.8,0.2); pen qqwwqq = rgb(0,0.4,0); /* draw figures */ draw(circle((3.45,1.27), 2.92), blue); draw((1.7,3.6)--(6.26,0.48), ccwwqq); draw((4.97,3.76)--(0.64,0.5), ccwwqq); draw((6.02,5.3)--(6.26,0.48), wwwwww); draw((0.4,5.32)--(0.64,0.5), wwwwww); draw((0.4,5.32)--(2.8,2.13), wwwwww); draw((0.64,0.5)--(2.42,3.1), wwwwww); draw((4.21,3.19)--(6.26,0.48), wwwwww); draw((0.4,5.32)--(4.97,3.76), wwwwww); draw((5.5,2.41)--(0.64,0.5), wwwwww); draw((6.02,5.3)--(1.7,3.6), wwwwww); draw((6.02,5.3)--(3.85,2.13), wwwwww); draw((4.97,3.76)--(4.96,0.48), wwwwww); draw((6.26,0.48)--(1.22,2.2), wwwwww); draw((0.49,3.53)--(6.09,3.82), ccwwqq); draw((0.64,0.5)--(6.26,0.48), ccwwqq); draw((6.26,0.48)--(4.58,4.73), ccwwqq); draw((0.4,5.32)--(6.26,0.48), linewidth(1.6) + ffcctt); draw((6.02,5.3)--(0.64,0.5), linewidth(1.6) + ffcctt); draw((1.7,3.6)--(4.96,2.2), linewidth(1.6) + ffcctt); draw((4.97,3.76)--(1.69,2.04), linewidth(1.6) + ffcctt); draw((4.58,4.73)--(1.68,0.5), linewidth(1.6) + qqwwqq); draw((0.49,3.53)--(5.5,2.41), linewidth(1.6) + qqwwqq); draw((1.22,2.2)--(6.09,3.82), linewidth(1.6) + qqwwqq); draw((1.7,3.6)--(1.68,0.5), wwwwww); draw((2.09,4.74)--(0.64,0.5), ccwwqq); draw((2.09,4.74)--(4.96,0.48), linewidth(1.6) + qqwwqq); draw(circle((1.17,2.05), 1.64), red); draw(circle((3.33,3.68), 1.64), red); draw(circle((5.61,2.12), 1.76), red); draw(shift((3.45,0.49))*xscale(2.81)*yscale(2.81)*arc((0,0),1,-0.2,179.8), red); /* dots and labels */ dot((1.7,3.6),dotstyle); label("$A$", (1.74,3.66), NE * labelscalefactor); dot((0.64,0.5),dotstyle); label("$B$", (0.68,0.56), NE * labelscalefactor); dot((6.26,0.48),dotstyle); label("$C$", (6.31,0.54), NE * labelscalefactor); dot((4.97,3.76),dotstyle); label("$D$", (5.01,3.82), NE * labelscalefactor); dot((3.85,2.13),dotstyle); label("$H_{DB}$", (3.9,2.19), NE * labelscalefactor); dot((4.96,0.48),dotstyle); label("$H_{DA}$", (5,0.55), NE * labelscalefactor); dot((1.69,2.04),dotstyle); label("$H_D$", (1.73,2.1), NE * labelscalefactor); dot((1.68,0.5),dotstyle); label("$H_{AD}$", (1.73,0.56), NE * labelscalefactor); dot((2.42,3.1),dotstyle); label("$H_{BD}$", (2.46,3.17), NE * labelscalefactor); dot((1.22,2.2),dotstyle); label("$H_{CD}$", (1.26,2.26), NE * labelscalefactor); dot((2.8,2.13),dotstyle); label("$H_{AC}$", (2.84,2.2), NE * labelscalefactor); dot((4.58,4.73),dotstyle); label("$H_{AB}$", (4.63,4.79), NE * labelscalefactor); dot((4.21,3.19),dotstyle); label("$H_{CA}$", (4.25,3.25), NE * labelscalefactor); dot((5.5,2.41),dotstyle); label("$H_{BA}$", (5.54,2.48), NE * labelscalefactor); dot((4.96,2.2),dotstyle); label("$H_{A}$", (5.01,2.27), NE * labelscalefactor); dot((0.49,3.53),dotstyle); label("$H_{BC}$", (0.53,3.6), NE * labelscalefactor); dot((6.09,3.82),dotstyle); label("$H_{CB}$", (6.13,3.88), NE * labelscalefactor); dot((6.02,5.3),dotstyle); label("$H_{B}$", (6.06,5.36), NE * labelscalefactor); dot((0.4,5.32),dotstyle); label("$H_{C}$", (0.44,5.38), NE * labelscalefactor); dot((3.33,2.9),dotstyle); dot((2.09,4.74),dotstyle); label("$H_{DC}$", (2.13,4.81), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]$

Proof: Denote by $H_{AB}$ the foot of the perpendicular from $A$ to the segment $CD$ (the one which does not contain $B$ ), and similarly define the $12$ other points of this type, as above. Denote by $H_A$ the orthocenter of $BCD$ , and define $H_B, H_C, H_D$ similarly. Observe that the red circles illustrated above with diameters $AB, BC, CD, DA$ contain each four other points.

From Pascal's Theorem on $AH_{AC}H_{AD}BH_{BC}H_{BD}$ , we get that $s_a, s_b, h_c$ are concurrent. Analogously, Pascal's Theorem on $AH_{AD}H_{AC}BH_{BD}H_{BC}$ gives that $s_a, s_b, h_d$ are concurrent. Similarly, all eight lines are concurrent, so done.