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

Labels:
median,
orthogonal,
projection,
triangle

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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