Proposed Problem
See complete Problem 263 at:
gogeometry.com/problem/p263_triangle_median_orthogonal_projection.htm
Level: High School, SAT Prep, College geometry
Wednesday, March 4, 2009
Problem 263: Triangle, Median, Altitude, Orthogonal Projection, Sides
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By cosine rule (Tr. ABC): a^2=b^2+c^2-2bccos(A)
ReplyDeletecos(A)=AH/AB = (b/2 - d)/c from Tr. BAH
Thus a^2 = b^2+c^2 -2bc*(b/2 - d)/c
= b^2+c^2 -2b*(b/2 - d)
= b^2 + c^2 -b^2 + 2bd
Thus, a^2 - c^2 = 2bd
Ajit
ajitathle@gmail.com
Solution using definition of Median and Pythagoras Theorem at http://www.osinfofrom.us/prob263.html
ReplyDeletex = AH = b/2 - d
ReplyDeletey = HC = b/2 + d
=> y - x = 2d
y + x = b
x^2 + BH^2 = c^2
y^2 + BH^2 = a^2
=> a^2 - c^2 = y^2 - x^2 = (y-x)(y+x) = 2db