Given a reference triangle △ABC, a reflection triangle △A′B′C′ is generated by reflecting each vertex about the opposite side, for example A′ is generated by reflecting vertex A about side BC. Now, suppose the vertices A′, B′ and C′ are given, how can you find the original vertices of the original triangle △ABC?
(In reply to
re(3): What am I missing? -> The Kosnita Point by broll)
Actually that mathworld picture gives a hint. The two triangles have the same orthocenter, H. The points A, H, A' are collinear and this line is perpendicular to BC.
So now find the point (A) on line HA' where the perpendiculars across HB' and HC' intersect HC' and HB' respectively at (C) and (B) and the line (B)(C) is perpendicular to HA'.
(Or we can solve the system implied by (2),(3),(4) to find the side lengths of ABC and go from there.)
|
|
Posted by Jer
on 2020-01-27 17:14:05 |
re(4): What am I missing? -> The Kosnita Point
