Construct any triangle ABC.
Construct the Orthocenter H of triangle ABC.
Construct the Orthocenter of triangle HBC.
Construct the Orthocenter of triangle HAB.
Construct the Orthocenter of triangle HAC.
1.The three hidden Orthocenters
I construct the Orthocenter H of triangle ABC, and secondly I construct the Orthocenter of triangle HBC, the Orthocenter of triangle HAB,and the Orthocenter of triangle HAC. But,I canot find the orthocenters, where wre they gone?
Conjectures
At first, Consider the triangle HAB, when we construct the Orthocenter of triangle HAB, we draw the perpendicular lines from H to AB, from A to BH,and from B to AH. Three lines intersect at the same point of vertex C, so C is conguruent to the Orthocenter of triangle HAB. In the same way, A is conguruent to the Orthocenter of triangle HBC, and B is conguruent to the Orthocenter of triangle HAC. That is why that the three Orthocenters are hidden.
Explanation of "The orthercenters of triangle HAB, HBC and HAC are conguruent to the Vertices C,A and B respectively."
To begin with, the orthercenter H of triangle ABC is the point
of intersection of three altitudes of triangle ABC, so it is obvious
that line AH is perpendicular to line BC, and line BH is perpendicular
to line AC, and line CH is perpendicular to line AB.(*)
Hence, the line that is perpendicular to AH through B is only
one and is conguruent to line BC. The line that is perpendicular
to BH through A is only one and is conguruent to line AC. The
line that is perpendicular to AB through H is only one and thruogh
the C. Hence, the orthercenter of triangle HAB is point C. The
otherorthercenters of triangle HBC, HCA are in the same way point
A and point B respectively.
2. Construct the Circumcircles of triangles ABC, HBC, HAB, and HAC. Conjectures?
Conjectures!
The length of radius of four circles are the same anyhow we
move original triangle ABC.
To prove this proposition, I am trying to prove following conjectures.
Quadrangle A(Hhbc) B(Hhac) Chbc Chac is parallelogram.
Quadrangle B(Hhac) C(Hhab) Chac Chac is parallelogram.
Quadrangle C(Hhab) A(Hhbc) Chac Chbc is parallelogram.
Quadrangle A(Hhbc) Chab H Chac is rhombus.
Quadrangle H Chbc ChabH Chac is rhombus.
Quadrangle Chab B(Hhac) Chbc H is rhombus.
3. Examine the triangle formed by the points where the extended altitudes meet the circumcircle. How is it related to theOrthic triangle?
I measured the lengths of sides of triangle FGH and sides of
triangle IJK, and the ratio of IK : FH and JI : GF and KJ : HG.
The ratio of IK : FH and JI : GF and KJ : HG are same 0.5.
It seems that the orthic triangle is similar to the triangle formed
by the extensions of those altitudes to the circumcircle, and
the ratio of similitude is 1 : 2.
4. Construct any acute triangle ABC and its circumcircle. Construct the three altitudes ha, hb, and hc. Extend each altitude to its intersection with the circumcircle and let Ha, Hb, and Hc be the corresponding segments from the vertex to the intersection with the circumcircle. Find!
The value of 
is always 4.00 any
how I move the triangle ABC.