perplexus.info :: Geometry : Cevian Theorem

Cevian Theorem (Posted on 2011-09-21)

Let AA', BB', CC', AB", and AC" be cevians of ΔABC with cevians AB" and AC" parallel to lines A'C' and A'B' respectively.

Prove that cevians AA', BB', and CC' are concurrent if and only if A' is the midpoint of line segment B"C".

Note: A cevian is a line segment which joins a vertex of a triangle with a point on the opposite side (or its extension).

Submitted by Bractals

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Solution: (Hide)

AB" || A'C' ==> ΔB"BA ~ ΔA'BC' ==> BA•BA' = BB"•BC' ==> BA•BA' = (BA' + A'B")•BC' ==> BA'•(BA - BC') = A'B"•BC' ==> BA'•C'A = A'B"•BC' ==> BA'•AC' = A'B"•C'B (1) AC" || A'B' ==> ΔC"CA ~ ΔA'CB' ==> CA•CA' = CC"•CB' ==> CA•CA' = (CA' + A'C")•CB' ==> CA'•(CA - CB') = A'C"•CB' ==> CA'•B'A = A'C"•CB' ==> -A'C•B'A = A'C"•CB' (2) Multiplying (1) & (2) together gives BA'•AC'•A'C"•CB' = -A'C•B'A•A'B"•C'B or BA'•CB'•AC'•A'C" = A'C•B'A•C'B•(-A'B") or BA' CB' AC' -A'B" -----•-----•----- = ------- (3) A'C B'A C'B A'C" Using (3) and Ceva's theorem Cevians AA', BB', and CC' are concurrent BA' CB' AC' <==> -----•-----•----- = 1 A'C B'A C'B -A'B" <==> ------- = 1 A'C" <==> A'B" = -A'C" <==> A' is the midpoint of B"C". QED

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	Subject	Author	Date
	Solution	Harry	2011-09-21 22:13:02

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