Thanks but I am having problems to show that ΔBEC is half of ΔABC

Please show atleast 1 and 5

P.S: Its already 10 o' clock here, so

Last edited by Agnishom (2013-03-13 03:39:49)

ABC and BDE are similar (both equilateral)

D is midpoint of BC => distances in BDE are half the equivalent distances in ABC.

So area = half base x height => area BDE is half area ABC  Q1

Also height of equilaterals are in same ratio

so comparing ABC and BCE

they have the same base BC and one has a height half the other => area BCE = half area ABC  Q3.

Still looking at the rest

Bob

Q2.

area ABE = half AB.BE.sin120 = half (2 BE)BE sin60 {sin 60 = sin 120}

= 2 (half BE.BE sin60) = 2 ar(BDE)

=> ar(BDE) = half ar(BAE)

Bob

Thanks but this is quite complicated.

We are not allowed to use trigonometry or similarity
The main properties we are supposed to use are:
1. Triangles on the same base and between same parallels have equal areas
2. A Median divides a triangle into two triangles of equal areas

hi

see diagram.

Construct midpoints L and M as shown.  DL ia parallel to AB etc.

Let BED have area = 3k

Using parallel rule the following also have area 3k

FMD, BDM, MLB, MLD, ALD, AML.  => LDC = 3k too.

ABC = 12k

ABD = ADC by median rule => ADC = 6k = ABD

ABF = ABD (parallel rule) => BFD = AFD

To do the rest I need to show that BF:FD = 2:1

At the moment I cannot see how to do that without using similarity so I'll post this and keep trying.

Bob

Uploaded Images

You are welcome. 

80/90 sounds good to me.  Let us know when you get the results.

And next time, allow a little more time before you post for help. 

Bob