# area of intersecting part of triangles

Let ABCD be a rectangle and AC one of its diagonals.Length of AD is 4m and of AB is 1 m.A line is drawn from point D towards the midpoint of BC which we will label E.Let DE intersect AC at F.How do we figure out the area of the triangle EFC?Any hint or link to previous solutions will be appreciated.

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Welcome to MSE! It helps if you share your thoughts and the things you've tried so we do not rehash those thoughts. Regards – Amzoti May 28 '13 at 18:16
That is the second time i have got that from you,funny thing.I just keep losing my accounts . – rah4927 May 28 '13 at 18:17

HINT: For $\text{h}$, you must know the length of $\text{EF}$ or $\text{FC}$ . Do you know in what ratio $\text{DE}$ cuts $\text{AC}$? Since $\triangle \text{AFD}$ and $\triangle \text{EFC}$ are similar, $\text{AF}=2\times\text{FC}$.

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I am afraid I do not know that. – rah4927 May 28 '13 at 18:20
@rah4927 have you been taught similar triangles? – user31280 May 28 '13 at 18:21
I was thinking about them a few seconds before you posted that comment.Thanks. – rah4927 May 28 '13 at 18:22

Hint You can easily show that $$A\left( \widehat { ABE } \right) =A\left( \widehat { AEC } \right) =\frac { 1 }{ 4 } A\left( ABCD \right)$$

because $E$ is midpoint. The most important part is to see $$AD=2EC\quad and\quad AD//EC$$

and therefore $$AF=2FC\quad and\quad A\left( \widehat { AEF } \right) =2A\left( \widehat { FEC } \right)$$

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