[alogo] 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.

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]
[2_0] [2_1] [2_2] [2_3]

[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.

See Also


Return to Gallery

Produced with EucliDraw©