# Calculating Perpendicular and Base of Triangle. Suggestion

In this diagram AB and CD are both perpendicular to BE.If EC=5 and CD=4. What is ratio of AB to BE ?

How would i go about solving this triangle (without trigonometric ratios). I could only get DE=3 using Pythagoras theorem and was stuck after that. How would i calculate BD ? Do i make BC ? Suggestions ?

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By Thales' theorem we have $\frac{CD}{AB}=\frac{DE}{BE}$ and so $\frac{AB}{BE}=\frac{CD}{DE}$.
We know $CD=4$ and by the Pythagorean theorem in $CDE$ we get $DE=3$.
So $\frac{AB}{BE}=\frac{4}{3}$.

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 Hi JBC thanks for the great answer. I am not really familiar with Thales theorem but i did read a little about. From What i could tell so far is , if we want to apply Thales on this figure it will be AC/CE = BD/DE. How did you come up with CD/AB = DE/BE – MistyD Jun 19 '12 at 22:37 Let $ABC$ a triangle, let $D\in(AB)$ and $E\in(AC)$ if $(DE)$ and $(BC)$ are parallel, then $\dfrac{AD}{AB}=\dfrac{AE}{AC}$ and $\dfrac{AD}{AB}=\dfrac{DE}{BC}$. See fr.wikipedia.org/wiki/Fichier:Thales_theorem_1.svg and fr.wikipedia.org/wiki/Fichier:Thales_theorem_2.svg for figures. – JBC Jun 19 '12 at 22:43

The two triangles $ABE$ and $CDE$ are similar. They have the same shape, You can think of $\triangle ABE$ as $\triangle CDE$, after you have put $\triangle CDE$ into a copier and scaled it up somewhat.

So the ratio of $AB$ to $BE$ is the same as the ratio of $CD$ to $DE$. But you calculated $DE$, so you know this ratio.

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@MistyD: Yes, that is correct. Easy, right? – André Nicolas Jun 19 '12 at 22:27
Yes. But I have some concerns : If they have he same shape does it necessarily mean they have the same ratio – MistyD Jun 19 '12 at 22:29
@MistyD: They are not merely both triangles. One is a scaled up version of the other, all angles match, so all ratios of dimensions are the same. – André Nicolas Jun 19 '12 at 22:31