Proposed Problem
Click the figure below to see the complete problem 305.
See also:
Complete Problem 305
Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Friday, June 19, 2009
Problem 305: Square, Triangles, Angle, Sides
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Triangles BCE, FDE and FAB are similar to each other, therefore, we can say that --> AB/BF = ED/FE. Since AB=BC, 1/BF = ED/(FE*BC). Further, BC/BE=FD/FE or 1/BE=FD/(FE*BC). Thus, (1/BE)^2 + (1/BF)^2 = [ED/(FE*BC)]^2+[FD/(FE*BC)]^2 =(1/BC)^2*{FE^2+ED^2)/FE^2 =(FE/FE)^2*(1/BC)^2= 1/BC^2.
ReplyDeleteHence the proof.
Let BC = a, BE = p, BF = q and DF = r
ReplyDeleteq/p = (a+r)/a
q^2 = a^2 + (a+r)^2
Substitute for a+r
q^2 = a^2 + a*2 q^2/p^2
Divide both sides by a^2 q^2
1/a^2 = 1/ p^2 + 1/q^2
Sumith Peiris
Moratuwa
Sri Lanka
Let angle EBC = theta = angle BFA. Then BC/BE = cos theta and BC/BF = sin theta, hence (BC/BE)^2 + (BC/BF)^2 = 1.
ReplyDeleteLet length of square ABCD be a & length of CE be b
ReplyDeleteED=a-b
Triangle BCE ~ Triangle FDE
BC/CE=FD/DE
a/b=FD/(a-b)
FD=a(a-b)/b
AF=a+a(a-b)/b=a^2/b
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BE^2=a^2+b^2
BF^2=a^2+a^4/b^2=(a^2b^2+a^4)/b^2
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1/BE^2+1/BF^2
=1/(a^2+b^2)+b^2/a^2*(a^2+b^2)
=1/a^2
=1/BC^2