Two Triangles With Common Base and Altitude

Two Triangles With Common Base and Altitude
What is this about?
A Mathematical Droodle

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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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