BitangentRectangularMember2

Rectangular hyperbola member

This is a continuation of the discussion started in BitangentRectangularMember.html . Here are some remarks concerning the location of the center of the rectangular hyperbola which is tangent to the sides {AB,AC} of triangle ABC at the vertices {B,C} correspondingly.
[1] The center O of the hyperbola is the projection of the orthocenter H on the median AA' of ABC.
[2] The intersection points {B₂,C₂} of the hyperbola with the circle on diameter BC are symmetric with respect to the center of the hyperbola.
[3] The other intersection points {B₃,C₃} of the hyperbola with the circumcircle are collinear with H', the middle of AH. Besides lines B₂C₂ and B₃C₃ are parallel.

[1] follows from the discussion in the aforementioned reference. There we saw that, to determine O, we construct a right-angled triangle similar to BCQ, Q being the intersection point of the circle with diameter BC with the median AA'. Then we form the right-angled triangles with the other sides of the triangle {BQB₁, CQC₁} and find the directions {B₀Q, C₀Q}. O is found on the median AA' by drawing parallels to those directions from B and C. Angle chasing shows that: BOC = (pi/2)+phi+psi = (3pi/2)-B₁-C₁ = (3pi/2)-(pi/2+A)=pi-A. But this is the angle by which H is viewing side BC, hence the result.
[2] follows from the remarks made in RectHypeParaChords.html , according to which, the other intersection points {B',C'} of the hyperbola with all circles passing through {B,C} define lines parallel to a fixed direction L₁, which is conjugate to the direction L₂, orthogonal to BC. Thus, all middles of chords B'C' (like B₂C₂, B₃C₃ etc.) are on line L₂, orthogonal to BC and passing through O.
Taking C₂ to be one of the intersection points of the circle with diameter BC, triangle BCC₂ is right angled, its orthocenter coincides with C₂ and the symmetric to O B₂ is the other intersection point with that circle. This proves [2] and determines line L₁.
To see [3] consider triangle BCC₃, where C₃ is one intersection point of the circumcircle with the hyperbola. The orthocenter H₀ of this triangle is also on the hyperbola and chord C₃H₀ is bisected at H₁ by L₁ (its conjugate diameter). Let now B₃ be the other intersection point with the circumcircle and M the middle of B₃C₃. Line H₁M, joins the middles and is parallel to B₃H₀ which has O for its middle. Hence H'H = C₃H₀/2 = DA', D being the circumcenter (bei well known property of the orthocenter, see f.e. Euler.html ). Since also DA' = AH/2 the proof is complete.
Note that AA₀, A₀ being an intersection point of the circumcircle with line L₂, is parallel and equal to HO. This can be seen by an argument similar to the previous one.

Remark These properties imply a construction of the hyperbola through five, out of the six {B,C,B₂,C₂,B₃,C₃} easily constructible points.

Rectangular hyperbola member

See Also