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AB':B'C = (s - a):(s - c)CA':A'B = s:(s - a)BC':C'A = (s - c): sProduct = 1So AA', BB', CC' are concurrent (by Ceva)
Is there a solution using only projective geometry?
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AB':B'C = (s - a):(s - c)
ReplyDeleteCA':A'B = s:(s - a)
BC':C'A = (s - c): s
Product = 1
So AA', BB', CC' are concurrent (by Ceva)
Is there a solution using only projective geometry?
ReplyDelete