Hello,

Do you know much about similar triangles ?

Since BD and AE are parallel, angles CBD and CAE are equal and angles CDB and CEA are equal.

angle ACE obviously equals angle BCD.

--> BCD and CAE are similar

Thus

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- September 7th 2008, 01:56 AM #1

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- September 7th 2008, 02:06 AM #2

- September 7th 2008, 02:27 AM #3

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- September 7th 2008, 02:31 AM #4
I'm dividing the length of the sides !

It's a property of similar triangles :

Similarity (geometry - Wikipedia, the free encyclopedia), see the second paragraph "similar triangles", there are equalities with the lengths of the sides.

- September 7th 2008, 03:04 AM #5

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Huh? :S

>_<

I'm very confused.

The wikipedia article does not explain**why/how**you divide**which**lengths of the triangle with each other, it just says that once you know everything congruent,**it is possible to****deduce****proportionalities between corresponding sides of the two triangles, such as the following**(and then list out a bunch).

Why are you dividing the lengths?

And how do you know which lengths to divide with each other?

And when you put the numbers in, is it like this?

6/(x+8)=6/(x+8)=x/(x+5)

If so, how would that work out?

>_<

- September 7th 2008, 03:25 AM #6
Okay.

See this table of corresponding angles and sides :

So it seems to be ok for the correspondance of the angles (you know which one equals which other).

Now, see what sides corresponding angles will intercept. This is what is said in the wikipedia.

After that, what wikipedia implicitly says is that the ratio of lengths of corresponding sides is a constant.

This is why I wrote the equalities above :

6/(x+8)=6/(x+8)=x/(x+5)

If so, how would that work out?

>_<

Well, see

This gives

Develop and solve the quadratic

- September 8th 2008, 04:42 PM #7

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