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Show that...

The picture says it all. "Vis at" means "show that". My first thought was that h is 2x, which is not correct. Maybe the formulas for area size is useful?

EDIT: (To make the question less dependent from the picture.)

A square with side $x$ is placed in the right triangle with legs $g$ and $h$. Show that $x=\frac{gh}{g+h}$.

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    I added the [trigonometry] tag.2011-08-22

3 Answers 3

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You can split the triangle into one triangle with base $g$ and height $x$, and another with base $h$ and height $x$. Just draw the diagonal line from the right angle to the opposite vertex of the $x\times x$ square. One of those triangles has area $gx/2$; the other has area $hx/2$. But they must add up to $gh/2$. Hence $ \frac{gx}{2} + \frac{hx}{2} = \frac{gh}{2}. $ By trivial algebra, the desired result will follow.

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From the geometry of the problem (see figure and identify two similar triangles (Wikipedia) or here), we can get the following proportions:

$\dfrac{h-x}{x}=\dfrac{x}{g-x}=\dfrac{h}{g}\tag{1}.$

enter image description here

$|AR|=h-x,|PR=x|,|PQ|=x,|BQ|=g-x,\measuredangle APR=\measuredangle PBQ,\measuredangle PAR=\measuredangle BPQ.$

Relation $(1)$ comprises 3 equations. Solve one of the them:

$\dfrac{h-x}{x}=\dfrac{x}{g-x},\tag{2}$

$\dfrac{h-x}{x}=\dfrac{h}{g},\tag{3}$

or

$\dfrac{x}{g-x}=\dfrac{h}{g}\tag{4}.$

  • Note 1: If it is required a proof using trigonometry (in the title), I can reformulate my answer. Added. If you apply trigonometry (mentioned in the title), note that $\tan \widehat{% ABC}=\frac{h}{g}$, $\tan \widehat{APR}=\frac{h-x}{x}$ and $\tan \widehat{APR}% =\tan \widehat{ABC}$. Use this equality and solve for $x$.

  • Note 2: If you haven't yet learned about similar triangles, I will modify my answer.

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    The questions were meant for the OP by the way to help with his/her understanding.2011-08-18
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Consider the area of the whole triangle and the areas of the constituent triangles and square.