## Some Perspectors From Mixed Perpendiculars

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• Dear All My Friends, Given triangle ABC with orthocenter H, circumcircle (O). We construct: Lc = perpendicular with AC passing through C Lb = perpendicular
Message 1 of 1 , Mar 5, 2007
Dear All My Friends,
Given triangle ABC with orthocenter H, circumcircle (O). We construct:
Lc = perpendicular with AC passing through C
Lb = perpendicular with AB passing through B
A1 = intersection of Lb, Lc
Ab = intersection of Lb, AH
Ac = intersection of Lc, AH
Similarly define B1, Bc, Ba, C1, Ca, Cb.
Easy to show that A1, B1, C1 are reflections of A, B, C resp. in O.
The results:
1. Three triangles A1AbAc, BaB1Bc, CaCbC1 are direct similar with ABC. Their circumcircles (Oa), (Ob), (Oc) are all orthogonal with circumcircle (O). Their circumradii Ra, Rb, Rc can be calculated as:
Ra = Abs(b^2 - c^2)*R/S
Rb = Abs(c^2 - a^2)*R/S
Rc = Abs(a^2 - b^2)*R/S
Here Abs(X) = absolute value of X and S =2*Area of ABC, R=circumradius of ABC.
So one of three values Ra, Rb, Rc is sum of other two.
2. ABC and OaObOc are perspective at symmedian point X(6).
3. OaObOc and median triangle of ABC are perspective at a point with barycentrics:
SA^2*(a^4 + (b^2 - c^2)^2) : :
Search value: +3.7554760336692 not in current ETC
4. OaObOc and orthic triangle of ABC are perspective at a point with barycentrics:
(a^4 + (b^2 - c^2)^2)/SA : :
Search value: -0.951003591246406 not in current ETC
Let three circles (Oa), (Ob), (Oc) cut circumcircle (O) at A2, B2, C2 resp. (other than A1, B1, C1).
5. ABC and A2B2C2 are perspective at X(25)
6. A2B2C2 and orthic triangle of ABC are perspective at centroid X(2)