Steiner line

Let $ABC$ be a triangle with orthocenter $H. P$ is a point on the circumcircle $\Omega$ of $\triangle ABC.$

Let $P_A, P_B,$ and $P_C$ be the reflections of $P$ in three lines which contains edges $BC, AC,$ and $AB,$ respectively.

Prove that $P_A, P_B, P_C,$ and $H$ are collinear. Respective line is known as the Steiner line of point $P$ with respect to $\triangle ABC.$

Proof

Let $D, E,$ and $F$ be the foots of the perpendiculars dropped from $P$ to lines $AB, AC,$ and $BC,$ respectively.

WLOG, Steiner line cross $AB$ at $Y$ and $AC$ at $Z.$

The line $DEF$ is Simson line of point $P$ with respect of $\triangle ABC.$

$D$ is midpoint of segment $PP_C \implies$ homothety centered at $P$ with ratio $2$ sends point $D$ to a point $P_C.$

Similarly, this homothety sends point $E$ to a point $P_B$ , point $F$ to a point $P_A,$ therefore this homothety send Simson line to line $P_AP_BP_C.$

Let $\angle ABC = \beta, \angle BFD = \varphi \implies \angle BDF = \beta – \varphi.$ $P_CP_A||DF \implies \angle P_CYB = \beta – \varphi.$ $P$ is simmetric to $P_C \implies \angle PYD = \beta – \varphi.$

Quadrungle $BDPF$ is cyclic $\implies \angle BPD = \varphi \implies \angle BPY = 90^\circ – \angle BYP – \angle BPD = 90^\circ – \beta.$

$\angle BCH = \angle BPY \implies PY \cap CH$ at point $H_C \in \Omega.$ Similarly, line $BH \cap PZ$ at $H_B \in \Omega.$

According the Collins Claim $YZ$ is $H-line,$ therefore $H \in P_AP_B.$

Simson line

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Collings Clime

Let triangle $ABC$ be the triangle with the orthocenter $H$ and circumcircle $\Omega.$ Denote $H–line$ any line containing point $H.$

Let $l_A, l_B,$ and $l_C$ be the reflections of $H-line$ in the edges $BC, AC,$ and $AB,$ respectively.

Prove that lines $l_A, l_B,$ and $l_C$ are concurrent and the point of concurrence lies on $\Omega.$

Proof

Let $D, E,$ and $F$ be the crosspoints of $H–line$ with $AB, AC,$ and $BC,$ respectively.

WLOG $D \in AB, E \in AC.$ Let $H_A, H_B,$ and $H_C$ be the points symmetric to $H$ with respect $BC, AC,$ and $AB,$ respectively.

Therefore $H_A \in l_A, H_B \in l_B, H_C \in l_C,$ $AH = AH_B = AH_C, BH = BH_A = BH_C, CH = CH_A = CH_B \implies$ $\angle HH_BE = \angle EHH_B = \angle BHD = \angle BH_CD.$

Let $P$ be the crosspoint of $l_B$ and $l_C \implies BH_CH_BP$ is cyclic $\implies P \in \Omega.$

Similarly $\angle CH_BE = \angle CHE = \angle CH_A \implies CH_BH_AP$ is cyclic $\implies P \in \Omega \implies$ the crosspoint of $l_B$ and $l_A$ is point $P.$

Usually the point $P$ is called the anti-Steiner point of the $H-line$ with respect to $\triangle ABC.$

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Ortholine

Let four lines made four triangles of a complete quadrilateral.

In the diagram these are $\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.$

Let points $H, H_A, H_B,$ and $H_C$ be the orthocenters of $\triangle ABC, \triangle ADE, \triangle BDF,$ and $\triangle CEF,$ respectively.

Prove that points $H, H_A, H_B,$ and $H_C$ are collinear.

Proof

Let $M$ be Miquel point of a complete quadrilateral.

Line $KLMN$ is the line which contain $4$ Simson lines of $4$ triangles.

Using homothety centered at $M$ with ratio $2$ we get $4$ coinciding Stainer lines which contain points $H, H_A, H_B,$ and $H_C$ .

Proof 2

$AH_A \perp DE, CH_C \perp EF \implies AH_A ||CH_C,$

$AH \perp BC, EH_C \perp CF \implies AH ||EH_C,$

$EH_A \perp AD, CH \perp AB \implies EH_A ||CH.$

Points $A, E,$ and $C$ are collinear.

According the Claim of parallel lines, points $H, H_A,$ and $H_C$ are collinear.

Similarly points $H, H_B,$ and $H_C$ are collinear as desired.

Claim of parallel lines

Let points $A, B,$ and $C$ be collinear.

Let points $D, E, F$ be such that $AF||CD, BF||CE, AE||BD.$

Prove that points $D, E,$ and $F$ are collinear.

Proof

Let $P = AE \cap CD, Q = AF \cap CE.$

$\angle CEP = \angle QEA, AQ||CP \implies \angle QAE = \angle CPE \implies$ $\triangle AEQ \sim \triangle PEC.$

$AP||BD \implies \frac {PD}{CD} = \frac {AB}{BC},$

$CQ||BF \implies \frac {AF}{QF} = \frac {AB}{BC} = \frac {PD}{CD}.$

The segments $EF$ and $ED$ are corresponding segments in similar triangles. Therefore $\angle CED = \angle QEF \implies D, E,$ and $F$ are collinear.

Complete Quadrilateral

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Shatunov-Tokarev line

Let the quadrilateral $ABCD$ be given ( $ABCD$ is not cyclic). Let points $E$ and $F$ be the midpoints of $BD$ and $AC,$ respectively. Let points $P$ and $Q$ be such points that $PA = PB, PC = PD, QA = QD, QB = QC.$

a) Prove that $PQ \perp EF.$

b) Prove that the point $X$ lies on the line $PQ$ iff $XA^2 + XC^2 = XB^2 + XD^2.$

Proof

a) Let $\omega$ be the circle centered at $F$ with radius $BE.$ Let $\Omega$ be the circle centered at $E$ with radius $AF.$ $PE$ is the median of $\triangle PBD \implies PE^2 = \frac {PB^2 + PD^2}{2} – BE^2.$

The power of the point $P$ with respect to the circle $\Omega$ is $Pow_{\Omega}(P) = PE^2 – AF^2 = \frac {PB^2 + PD^2}{2} – BE^2 – AF^2.$

$PF$ is the median of $\triangle PAC \implies PF^2 = \frac {PA^2 + PC^2}{2} – AF^2.$

The power of the point $P$ with respect to the circle $\omega$ is $Pow_{\omega}(P) = PF^2 – BE^2 = \frac {PA^2 + PC^2}{2} – BE^2 – AF^2 = \frac {PB^2 + PD^2}{2} – BE^2 – AF^2 = Pow_{\Omega}(P).$

Therefore $P$ lies on the radical axis of $\Omega$ and $\omega.$ Similarly, $Q$ lies on these line. So the line $PQ$ is the radical axes of $\Omega$ and $\omega.$

This line is perpendicular to Gauss line $EF$ which is the line of centers of two circles $\Omega$ and $\omega$ as desired.

b) $XE$ is the median of $\triangle XBD \implies XE^2 = \frac {XB^2 + XD^2}{2} – BE^2.$

$XF$ is the median of $\triangle XAC \implies XF^2 = \frac {XA^2 + XC^2}{2} – AF^2.$

$X$ lies on the radical axes of $\Omega$ and $\omega \implies XE^2 – XF^2 = AF^2 – BE^2 \implies$ $\frac {XB^2 + XD^2}{2} – BE^2 – ( \frac {XA^2 + XC^2}{2} – AF^2) = AF^2 – BE^2 \implies XB^2 + XD^2 = XA^2 + XC^2.$

If the point $X$ satisfies the equation $XB^2 + XD^2 = XA^2 + XC^2$ then locus of $X$ is the straight line (one can prove it using method of coordinates).

The points $P$ and $Q$ are satisfies this equation, so this line contain these points as desired.

It is easy to understand that this line is parallel to Steiner line which is the radical axis of the circles centered at $E$ and $F$ with radii $BE$ and $AF,$ respectively.

Of course, it is parallel to Simson line.

Complete Quadrilateral

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Shatunov-Tokarev concurrent lines

Let the quadrilateral $ABCD$ be given ( $ABCD$ is not cyclic).

Let points $A'$ and $A''$ be on the line $AB$ such that $AA' = AA''$ . Similarly $B' \in BC, B'' \in BC, C' \in CD, C'' \in CD,$ $D' \in AD, D'' \in AD,$ $BB' = BB'' = CC' = CC'' = DD' = DD'' = AA'.$

Let points $Q, Q',$ and $Q''$ be the crosspoints of the bisectors $AD \cap BC, A'D' \cap B'C', A''D'' \cap B''C''.$

Similarly points $P, P',$ and $P''$ are the crosspoints of the bisectors $AB \cap CD, A'B' \cap C'D', A''B'' \cap C''D''.$

Prove that lines $PQ, P'Q',$ and $P''Q''$ are concurrent.

Proof

Segment $XA$ is the median of the $\triangle XA'A'' \implies 2(XA^2 + AA'^2) = XA'^2 + XA''^2.$

Similarly $2(XB^2 + BB'^2) = XB'^2 + XB''^2, 2(XC^2 + CC'^2) = XC'^2 + XC''^2, 2(XD^2 + DD'^2) = XD'^2 + XD''^2.$

Let $PQ$ cross $P'Q'$ at point $X \implies XA^2 + XC^2 = XB^2 + XD^2, XA'^2 + XC'^2 = XB'^2 + XD'^2.$

We made simple calculations and get $XA''^2 + XC''^2 = XB''^2 + XD''^2,$ therefore point $X$ lies on $P''Q''$ as desired. vladimir.shelomovskii@gmail.com, vvsss

Shatunov point

Let the quadrilateral $ABCD$ be given ( $ABCD$ is not cyclic).

Let points $A', B', C',$ and $D'$ be on the lines $AD, AB, BC,$ and $CD,$ respectively such that $|AA'| = |BB'| = |CC'| = |DD'| = d.$

Let points $A'', B'', C'',$ and $D''$ be on the segments $AA', BB', CC',$ and $DD',$ respectively such that $\frac {|AA''|}{|A'A''|} = \frac {|BB''|}{|B'B''|} = \frac {|CC''|}{|C'C''|} = \frac {|DD''|}{|D'D''|} = \frac {m}{n},$ where $m + n = 1.$

Let points $Q, Q',$ and $Q''$ be the crosspoints of the bisectors $AD \cap BC, A'D' \cap B'C', A''D'' \cap B''C''.$

Similarly points $P, P',$ and $P''$ are the crosspoints of the bisectors $AB \cap CD, A'B' \cap C'D', A''B'' \cap C''D''.$

Prove that lines $PQ, P'Q',$ and $P''Q''$ are concurrent.

Proof

Segment $XA''$ is the cevian to the side AA' of the $\triangle XAA'.$

We use the Stewart's theorem and get: $m \cdot|XA'|^2 + n \cdot |XA|^2) = |XA''|^2 + mn \cdot d^2.$ Similarly $m \cdot|XB'|^2 + n \cdot |XB|^2) = |XB''|^2 + mn \cdot d^2,$ $m \cdot|XC'|^2 + n \cdot |XC|^2) = |XC''|^2 + mn \cdot d^2,$ $m \cdot|XD'|^2 + n \cdot |XD|^2) = |XD''|^2 + mn \cdot d^2.$ Let $PQ$ cross $P'Q'$ at point $X \implies |XA|^2 + |XC|^2 = |XB|^2 + |XD|^2, |XA'|^2 + |XC'|^2 = |XB'|^2 + |XD'|^2.$