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Area of triangles questions

Posted March 4th, 2008 by alpha
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An equilateral triangle ABC is given. D is the mid-point of BC. Another equilateral triangle BDE is formed and its vertex E lies opposite to A. Line AE intersects BC at F.

Prove that-
1)area of BFE= 2 area of FED
2)area of FED = 1/8 area of AFC

‹ Find the coordinates of the centre of a circle passing through the given points. Prove that the areas are equal ›

Comments

I do not understand your 4th

On March 11th, 2008 alpha says:

I do not understand your 4th and 5th steps(2BF=FC,
2DF=FB).Can you further prove those steps?

answer

On March 11th, 2008 Structure says:

Let AB=AC=BC=6a
Then BD=DE=EB=3a
EB parallel to AC

$$\frac{AF}{EF}=\frac{FC}{FB}=\frac{CA}{BE}=\frac{6a}{3a}=2$$

2EB=AC
From the second quote we have
2BF=FC but BC=BF+FC=BF+2BF=3BF=6a
BF=2a
BD=3a BF=2a so

$$FD=a$$

2DF=FB
Triangles having the same altitude EFB and EDF have area proportional to basis
Then 2Area(EDF)=Area(EFB)
Similar triangles ACF and EFB have similarity quote 2 then area quote 4 so
area (ACF)=8 area (FED)

Area of similar triangles

On March 7th, 2008 Structure says:

EB parallel to AC
2EB=AC
2BF=FC
2DF=FB
Triangles having the same altitude EFB and EDF have area proportional to basis
Then 2Area(EDF)=Area(EFB)
Similar triangles ACF and EFB have similarity quote 2 then area quote 4 so
area (ACF)=8 area (FED)

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