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

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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 ?

Edit: The answer is 4:3.

2 Answers 2

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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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    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 http://fr.wikipedia.org/wiki/Fichier:Thales_theorem_1.svg and http://fr.wikipedia.org/wiki/Fichier:Thales_theorem_2.svg for figures.2012-06-19
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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: They are $n$ot merely both triangles. One is a scaled up version of the other, all angles match, so **all** ratios of dimensions are the same.2012-06-19