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
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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.
See the drawing
ReplyDeleteIn ΔBHC : ∠ABC + ∠HCB =90 °
In ΔBCA : ∠HCA + ∠HCB =90° => ∠HCA = ∠ABC=β
In ΔCHA : ∠HCA +∠BAC=90° => ∠HCB =∠BAC= α
=> ΔBHC is similar to ΔCHA (aaa=β90α)
and ΔBHC is similar to ΔBCA (aaa=β90α)
Therefore ΔBHC, ΔCHA and ΔBCA are similar
c=m+n
(1) In ΔBHC : a^2 = m^2+h^2
(2) In ΔCHA : b^2 = h^2+n^2
(3) In ΔBCA : a^2+b^2 = (m+n)^2
(4=1+2) a^2+b^2 = m^2+n^2+2h^2
(5=4+3) m^2+n^2+2h^2 = (m+n)^2 = m^2+n^2+2mn => h^2= mn
(1+5): a^2 = m^2+h^2=m^2+mn =m(m+n)=> a^2=cm
(2+5): b^2 = n^2+h^2=n^2+mn=n(m+n)=> b^2=cn