Midpoint Areas (Posted on 2012-02-18)

	Submitted by Bractals
Rating: 4.0000 (1 votes)
Solution:	(Hide)
NOTATION:    PQ denotes the vector from point P to point Q.    ⊗ denotes the vector cross product. PROOF: Let BD = xDC, CE = yEA, and AF = zFB. From Ceva's theorem we know that lines    AD, BE, and CF are concurrent               ⇔             xyz = 1.                  (1)   Triangles BDF and CED have the   same orientation and equal areas  ⇔ BD⊗BF = CE⊗CD  ⇔ [x/(1+x)](AC-AB)⊗[-1/(1+z)]AB =     [-y/(1+y)]AC⊗[-1/(1+x)](AC-AB)  ⇔ [-x/((1+x)(1+z))]AC⊗AB =     [-y/((1+x)(1+y))]AC⊗AB  ⇔ x/(1+z) = y/((1+y)                (2) Both sets of equations (3) and (4) satisfy equations (1) and (2):    x = 1 and z = 1/y                  (3)    x = -1/(1+y) and z = -(1+y)/y      (4) The set of equations (4) imply    CF = (1+y)AB - AC    BE = -1/(1+y)CF    AD = 1/yCF In other words, lines AD, BE, and CF are parallel. Thus,   Triangles BDF and CED have the   same orientation and equal areas  ⇔ x = 1  ⇔ BD = DC  ⇔ D is the midpoint of side BC. QED

	Subject	Author	Date
	re: Possible solution	Bractals	2012-02-19 14:07:11
	Possible solution	broll	2012-02-19 08:02:30