Proposed Problem

Click the figure below to see the complete problem 305.

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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