Good Afternoon!!!
Please, can you help me with it?
Where AB = 5, AD = 3, CE = 1. The area of the triangle BCF is?
I started making some analogy between similar triangles:
First: between ABF and CEF
But Im not sure...
What can I do now?!
Yes, you can continue along that route if you like.
Triangles ABF and CEF are similar. ABF is CEF magnified.
The base of ABF is 5 and the base of CEF is 1.
This means that the perpendicular height of CEF is one-fifth of the perpendicular height of ABF,
since the linear measurements of ABF are 5 times those of CEF.
Hence you must divide AD or BC into 6 equal subdivisions.
Then you have the base and perpendicular height of CEF,
so you can find it's area.
Can you finish ?
OOps!!! sorry, that's ECF not BCF.
However, you then only need subtract the area of ECF from ECB which is easy.
Ahhhh! Now I undertood!
I made the 'divison' as you said (but of the BCF!) , and I fond the height!
$\displaystyle h=\displaystyle{\frac{5}{6}}$
Then I just calculate the area!
$\displaystyle \displaystyle{\frac{BC.h}{2} = \displaystyle{\frac{5}{4}$
Thanx!!!!