Difference between revisions of "Complete Quadrilateral"
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*[[Simson line]] | *[[Simson line]] | ||
*[[Steiner line]] | *[[Steiner line]] | ||
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| + | ==Radical axis== | ||
| + | [[File:Complete radical axes.png|400px|right]] | ||
| + | Let four lines made four triangles of a complete quadrilateral. In the diagram these are <math>\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.</math> | ||
| + | |||
| + | Let points <math>H,</math> and <math>H_A</math> be the orthocenters of <math>\triangle ABC</math> and <math>\triangle ADE,</math> respectively. | ||
| + | |||
| + | Let circles <math>\omega, \theta,</math> and <math>\Omega</math> be the circles with diameters <math>CD, BE,</math> and <math>AF,</math> respectively. | ||
| + | Prove that Steiner line <math>HH_A</math> is the radical axis of <math>\omega, \theta,</math> and <math>\Omega.</math> | ||
| + | |||
| + | <i><b>Proof</b></i> | ||
| + | |||
| + | Let points <math>G, K, L, N, P,</math> and <math>Q</math> be the foots of perpendiculars <math>AH_A, CH, DH_A, BH, AH,</math> and <math>EH_A,</math> respectively. | ||
| + | |||
| + | Denote <math>Po(X)_{\omega}</math> power of point <math>X</math> with respect the circle <math>\omega.</math> | ||
| + | <cmath>\angle AGF = 90^\circ \implies G \in \Omega \implies Po(H_A)_{\Omega} = AH_A \cdot GH_A.</cmath> | ||
| + | <cmath>\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}.</cmath> | ||
| + | <cmath>\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}.</cmath> | ||
| + | <cmath>\angle APF = 90^\circ \implies P \in \Omega \implies Po(H)_{\Omega} = AH \cdot PH_A.</cmath> | ||
| + | <cmath>\angle BNE = 90^\circ \implies N \in \theta \implies Po(H)_{\theta} = BH \cdot NH = AH \cdot PH = Po(H)_{\Omega}.</cmath> | ||
| + | <cmath>\angle CKD = 90^\circ \implies K \in \theta \implies Po(H)_{\omega} = CH \cdot KH = AH \cdot PH = Po(H)_{\Omega}.</cmath> | ||
| + | Therefore power of points <math>H</math> and <math>H_A</math> with respect these three circles are the same, these points lies on the common radical axis of <math>\omega, \theta,</math> and <math>\Omega \implies</math> Steiner line <math>HH_A</math> is the radical axis as desired. | ||
Revision as of 15:31, 9 December 2022
Complete quadrilateral
Let four lines made four triangles of a complete quadrilateral. In the diagram these are 
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
Let points 
and 
be the orthocenters of 
and 
respectively.
Let circles 
and 
be the circles with diameters 
and 
respectively.
Prove that Steiner line 
is the radical axis of and 
Proof
Let points 
and 
be the foots of perpendiculars 
and 
respectively.
Denote 
power of point 
with respect the circle 
![\[\angle AGF = 90^\circ \implies G \in \Omega \implies Po(H_A)_{\Omega} = AH_A \cdot GH_A.\]](http://latex.artofproblemsolving.com/1/7/5/1751f7708cd7afeb2a2eba5cc0fedac7cb644fcb.png)
![\[\angle CLD = 90^\circ \implies L \in \omega \implies Po(H_A)_{\omega} = DH_A \cdot LH_A = AH_A \cdot GH_A = Po(H_A)_{\Omega}.\]](http://latex.artofproblemsolving.com/b/6/d/b6d42cf00a5e05edc26bc604bcf902fa06a9047f.png)
![\[\angle EQB = 90^\circ \implies Q \in \theta \implies Po(H_A)_{\theta} = EH_A \cdot QH_A = AH_A \cdot GH_A = Po(H_A)_{\Omega}.\]](http://latex.artofproblemsolving.com/6/e/e/6eedb5b28f2974f98beb1844d2b895f202111a3a.png)
![\[\angle APF = 90^\circ \implies P \in \Omega \implies Po(H)_{\Omega} = AH \cdot PH_A.\]](http://latex.artofproblemsolving.com/8/f/4/8f42a140addc160d20dd07242fe30414d9f4086f.png)
![\[\angle BNE = 90^\circ \implies N \in \theta \implies Po(H)_{\theta} = BH \cdot NH = AH \cdot PH = Po(H)_{\Omega}.\]](http://latex.artofproblemsolving.com/5/9/c/59c3e66a0e8c6bb2d7283fd28a0b8cc605fb00ed.png)
![\[\angle CKD = 90^\circ \implies K \in \theta \implies Po(H)_{\omega} = CH \cdot KH = AH \cdot PH = Po(H)_{\Omega}.\]](http://latex.artofproblemsolving.com/5/e/d/5edbb58fd034495eff113e3818b948a53f5f4459.png)
Therefore power of points 
and with respect these three circles are the same, these points lies on the common radical axis of and 
Steiner line is the radical axis as desired.
