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Two Triangles With Common Base and Altitude: What is this about?
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The applet may suggest a statement by Greg Markowsky. If C and D are two points on a semicircle with diameter AB, E is the intersection of AD and BC and F the foot of perpendicular from E to AB, then Greg calls C and D a pair of reflection about F based on an observation of his.
| Given a semicircle on diameter AB and four points C, D, U, V, on the semicircle. Assume the points form two pairs of reflection - {C, D} and {U, V} - about a point F. Then DU and CV intersect on the perpendicular to AB at F. |
Greg's statement admits a reformulation: Given two triangles ABS and ABT with altitudes AD, BC, FS and AV, BU, FT, respectively. Then DU and CV intersect on the common altitude through F.
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Proof
Consider the hexagon ADUBCV and a perpendicular FG to AB. By the mirror property of altitudes AD and BC cross on FG and so are VA and UB. Let X be the intersection of DU and CV. By Pascal's Hexagram Theorem, the three points of intersection are collinear. Since the first two lie on FG, so does X.
References
- G. Markowsky, Pascal’s Hexagon Theorem implies the Butterfly Theorem, 2007, submitted for publication
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