Collinear Feet (Posted on 2010-09-26)

	Submitted by Bractals
Rating: 3.0000 (2 votes)
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Let H be the orthocenter and P, Q, R, and S the feet of the perpendiculars from A' to AB, BB', CC', and AC respectively. Construct the line segments PQ, QR, and RS. -------------------------- ------------- Quadrilaterals A'BPQ and A'QHR are cyclic since side A'B subtends equal angles at P and Q and angles A'QH and A'RH are supplementary. Therefore,      /BQP = /BA'P = /BCC' = /A'CH = 90° - /A'HC                = 90° - /A'HR = /HA'R = /HQR Thus, P, Q, and R are collinear. ------------- Repeat the above argument with the following letters (B-C, P-S, and Q-R) swapped and we have S, R, and Q are collinear. Therefore, P, Q, R, and S are collinear. -------------------------- Note: The proposition holds for all triangles.           The proof is trivial if A, B, or C is a right angle.           The proof when A is obtuse is the same as above           with the following letters (A-H, B'-C', P-Q, and           R-S) swapped. The proof when C is obtuse is the           same as the proof when B is obtuse with the           following letters (B-C, B'-C', P-S, and Q-R)           swapped. The proof for when B is obtuse is           left to the reader.

	Subject	Author	Date
	Possible spoiler	broll	2010-09-26 12:07:58