Complete Quadrilateral

Complete quadrilateral

Let four lines made four triangles of a complete quadrilateral. In the diagram these are $\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.$ One can see some of the properties of this configuration and their proof using the following links.

Radical axis

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 $H,$ and $H_A$ be the orthocenters of $\triangle ABC$ and $\triangle ADE,$ respectively.

Let circles $\omega, \theta,$ and $\Omega$ be the circles with diameters $CD, BE,$ and $AF,$ respectively. Prove that Steiner line $HH_A$ is the radical axis of $\omega, \theta,$ and $\Omega.$

Proof

Let points $G, K, L, N, P,$ and $Q$ be the foots of perpendiculars $AH_A, CH, DH_A, BH, AH,$ and $EH_A,$ respectively.

Denote $Po(X)_{\omega}$ power of point $X$ with respect the circle $\omega.$ $\angle AGF = 90^\circ \implies G \in \Omega \implies Po(H_A)_{\Omega} = AH_A \cdot GH_A.$ $\angle APF = 90^\circ \implies P \in \Omega \implies Po(H)_{\Omega} = AH \cdot PH_A.$ $\angle CLD = 90^\circ \implies L \in \omega \implies Po(H_A)_{\omega} = DH_A \cdot LH_A = AH_A \cdot GH_A = Po(H_A)_{\Omega}.$ $\angle EQB = 90^\circ \implies Q \in \theta \implies Po(H_A)_{\theta} = EH_A \cdot QH_A = AH_A \cdot GH_A = Po(H_A)_{\Omega}.$

$\angle BNE = 90^\circ \implies N \in \theta \implies Po(H)_{\theta} = BH \cdot NH = AH \cdot PH = Po(H)_{\Omega}.$ $\angle CKD = 90^\circ \implies K \in \theta \implies Po(H)_{\omega} = CH \cdot KH = AH \cdot PH = Po(H)_{\Omega}.$ Therefore power of point $H (H_A)$ with respect these three circles is the same. These points lies on the common radical axis of $\omega, \theta,$ and $\Omega \implies$ Steiner line $HH_A$ is the radical axis as desired.

vladimir.shelomovskii@gmail.com, vvsss

Areas in complete quadrilateral

Let complete quadrilateral $ABCDEF$ be given $(E = AC \cap BD, F = AB \cap CD)$ . Let $X, Y,$ and $Z$ be the midpoints of $BC, AD,$ and $EF,$ respectively.

Prove that $\frac {[ADFE]}{[ADC]} = \frac {ZY}{XY},$ where $[t]$ is the area of $t.$

Proof

Let $X_1, Y_1, D_1, C_1, Z_1,$ and $F_1$ be the projections of $X, Y, A, B, Z,$ and $E,$ respectively onto $CD.$ Let $\omega, \theta,$ and $\Omega$ be the circles with diameters $AD, BC,$ and $EF,$ respectively. $\angle EF_1F = 90^\circ \implies F_1 \in \Omega.$ Similarly $D_1 \in \omega, C_1 \in \theta.$ Let common radical axes of $\omega, \Omega,$ and $\theta$ cross $CD$ at point $G.$ $EF_1 || AD_1 \implies \frac {EC}{F_1C} = \frac {AC}{D_1C}.$ The power of the point $G$ with respect $\omega, theta,$ and $\Omega$ is the same, therefore $GF \cdot GF_1 = GC \cdot GC_1 = GD \cdot GD_1 \implies$ $(G - F) \cdot (G - F_1) = (G - C) \cdot (G - C_1) = (G - D) \cdot (G – D_1) \implies$ $F \cdot F_1 – G \cdot (F + F_1) = F \cdot F_1 – G \cdot 2Z_1 = C \cdot C_1 – G \cdot (C + C_1) =$ $$ (Error compiling LaTeX. Unknown error_msg)= C \cdot C_1 – G \cdot 2X_1 = D \cdot D_1 – G \cdot (D + D_1) = D \cdot D_1 – G \cdot 2Y_1 \implies$$ (Error compiling LaTeX. Unknown error_msg)G = \frac {F \cdot F_1 – C \cdot C_1}{2(Z_1 - X_1)} = \frac {F \cdot F_1 – D \cdot D_1}{2(Z_1 – Y_1)} \implies$$ (Error compiling LaTeX. Unknown error_msg)\frac {Z_1 - X_1}{Z_1 - Y_1} = \frac {F \cdot F_1 – C \cdot C_1} {F \cdot F_1 – D \cdot D_1}$$ (Error compiling LaTeX. Unknown error_msg)\frac {Z - X}{Z - Y} = \frac {ZX}{ZY} = \frac {FC \cdot F_1C} {DC \cdot D_1C} = \frac {FC \cdot DC} {EC \cdot BC} = \frac {[CEF]}{[CAD]}.$ Therefore $\frac {[ADEF]}{[ACD]} = \frac {[CEF]-[CAD]}{[CAD]} =\frac {ZX - ZY}{ZY} = \frac {XY}{ZY}.$

vladimir.shelomovskii@gmail.com, vvsss

Newton–Gauss line

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 $K, L,$ and $N$ be the midpoints of $BE, CD,$ and $AF,$ respectively.

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

Prove that Steiner line $HH_A$ is perpendicular to Gauss line $KLN.$

Proof

Points $K, L,$ and $N$ are the centers of circles with diameters $BE, CD,$ and $AF,$ respectively.

Steiner line $HH_A$ is the radical axis of these circles.

Therefore $HH_A \perp KL$ as desired.

vladimir.shelomovskii@gmail.com, vvsss

Shatunov-Tokarev line

Let the complete quadrilateral ABCDEF be labeled as in the diagram. Quadrilateral $BCED$ is not cyclic.

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

Let bisector $BD$ cross bisector $CE$ at point $Q.$ Let bisector $BC$ cross bisector $DE$ at point $P.$

Prove that

a) points $P$ and $Q$ lie on circumcircle of $\triangle OO_AO_C,$

b) line $PQ$ is symmetric to Steiner line with respect centroid of $BDEC.$

I suppose that this line was found independently by two young mathematicians Leonid Shatunov and Alexander Tokarev in 2022. I would be grateful for information on whether this line was previously known.

Proof

a) Points $P$ and $O$ lies on bisector of $BC,$ points $P$ and $O_A$ lies on bisector of $DE \implies$

$\angle OPO_A + \angle OO_CO_A = 180^\circ \implies P \in$ circle $OO_AO_C.$

Similarly $P \in$ circle $OO_BO_C$ as desired.

Miquel's point

b) Let $M, M', P'$ and $G$ be midpoints of $CE, BD, HH_A,$ and $MM',$ respectively.

It is clear that $G$ is centroid of $BDEC.$

$M'P \perp BD, EH_A \perp AD, CH \perp AB \implies MP' \perp AB$ (midline of trapezium $ECHH_A) \implies EH_A||MP'||CH||M'P.$ $MP \perp CE, DH_A \perp AE, BH \perp AC \implies$ $M'P' \perp AB$ (midline of trapezium $HBDH_A) \implies$ $DH_A||M'P'||BH||MP \implies M'P'MP$ is parallelogram.