Wednesday, October 13, 2010

Problem 528: Triangle, Medians, Perpendicular, Measurement

Geometry Problem
Click the figure below to see the complete problem 528 about Triangle, Medians, Perpendicular, Measurement.

 Problem 528: Triangle, Medians, Perpendicular, Measurement.
See also:
Complete Problem 528

Level: High School, SAT Prep, College geometry

1 comment:

  1. Solution sent by Peter Tran:

    Let G is the intersection of AD and CE
    We have AD^2= ½*(AB^2+AC^2-BC^2/2) …….. Formula to calculate the median
    AG= 2/3*AD .
    Replace AC=b , AB=c and BC=a we get AG^2=2/9*(c^2+b^2-a^2/2)
    Similarly we also have CG^2=2/9*(a^2b^2-C^2/2)
    But AC^2=AG^2+CG^2 .
    Replace values of AG^2 and CG^2 from above formulas we will get a^2+c^2=5.b^2
    Peter Tran