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- Sep 7th 2008, 12:56 AM #1

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- Sep 7th 2008, 01:06 AM #2
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 $\displaystyle \frac{BC}{AC}=\frac{CD}{CE}=\frac{BD}{AE}$

- Sep 7th 2008, 01:27 AM #3

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- Sep 7th 2008, 01:31 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.

- Sep 7th 2008, 02:04 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?

>_<

- Sep 7th 2008, 02:25 AM #6
Okay.

See this table of corresponding angles and sides :

$\displaystyle \begin{array}{|c|c|} \hline \text{Triangle BCD} & \text{Triangle CAE} \\ \hline \angle BCD & \angle ACE \\ \hline \angle CBD & \angle CAE \\ \hline \angle CDB & \angle CEA \\ \hline \hline \hline CB & CA \\ \hline CD & CE \\ \hline BD & AE \\ \hline \end{array}$

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 :

$\displaystyle \frac{BC}{AC}=\frac{CD}{CE}=\frac{BD}{AE}

$

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

If so, how would that work out?

>_<

Well, see $\displaystyle \frac{6}{x+8}=\frac{x}{x+5}$

This gives $\displaystyle 6(x+5)=x(x+8)$

Develop and solve the quadratic

- Sep 8th 2008, 03:42 PM #7

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