Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 749 details.

## Friday, May 4, 2012

### Problem 749: Triangle, Rhombus, Parallel, Harmonic Means, Metric Relations

Labels:
harmonic mean,
parallel,
rhombus,
triangle

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Since ABEF is a rhombus so AE is an angle bisector of angle BAC

ReplyDeleteWe have BE/EC=AB/AC= c/b

BE/BC= c/(b+c)=x/b => x=b.c/(b+c)

ADEF is a rhombus.FE = x = DE

ReplyDeletex/b = DE/AC = BE/BC and

x/c = FE/AB = CE/BC

Add:

x/b + x/c = (BE + CE)/BC = BC/BC = 1

x(1/b + 1/c)= 1

x = bc /(b+c)

Also

x = (1/2)[2/(1/b + 1/c)]:=(1/2)HM(b,c)

Tr DBE ~ tr EFC => x / (b+x) = (c-x) / x

ReplyDelete