Thursday, March 5, 2009

Problem 264: Right Triangle, Altitude, Leg projection, Hypotenuse, Similarity, Geometric mean

Proposed Problem
Problem 264. Right Triangle, Altitude, Leg projection, Hypotenuse, Similarity, Geometric mean.

See complete Problem 264 at:
gogeometry.com/problem/p264_right_triangle_leg_projection_hypotenuse.htm

Level: High School, SAT Prep, College geometry

2 comments:

  1. ang BCH =an CAH =an CAB ( CB perpendicular to AC,& CH to
    AB)
    => tr BHC~ tr CHA ~ tr BCA

    BCH ~ BCA => BC/BA = m/a => a/c = m/a => a'2 = m*c

    BCA ~ CAH => b/c = h/b => b'2 = c*b

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  2. Let angle CHA=a and angle CBH=b...(1)
    tri BCH and tri CAH are right angled triangles.....(2)
    by (1) and (2) and T(sum of the measures of all interior angles is 180),
    angle BCH=a and angle HCA=b...(3)
    Hence, by (2) and (3) and AAA test of similarity,
    tri BHC ~ tri CHA ~ tri BCA.
    h^2 = mn...(4)
    In tri CHA by pythagoras theorem,
    CH^2 + HA^2 =CA^2
    hence,n^2 + h^2 = b^2
    n^2 + mn = b^2
    n (n + m) = b^2
    n * c = b^2
    In tri BHC by pythagoras theorem,
    CH^2 + HB^2 =CB^2
    hence,m^2 + h^2 = c^2
    m^2 + mn = c^2
    m (n + m) = c^2
    m * c = c^2.

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