Proposed Problem

See complete Problem 264 at:

gogeometry.com/problem/p264_right_triangle_leg_projection_hypotenuse.htm

Level: High School, SAT Prep, College geometry

## Thursday, March 5, 2009

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

Labels:
cathetus,
orthogonal,
projection,
right triangle,
similarity

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ang BCH =an CAH =an CAB ( CB perpendicular to AC,& CH to

ReplyDeleteAB)

=> 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

Let angle CHA=a and angle CBH=b...(1)

ReplyDeletetri 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.