# Algebra in trigonometry, algebraic proof?

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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Do you require a proof using trigonometry? If not, the title should be revised. – Américo Tavares Aug 17 '11 at 15:53
I added the [trigonometry] tag. – Américo Tavares Aug 22 '11 at 11:03

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:

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

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

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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If you noticed this is a homework problem, so perhaps the OP's teachers wanted them to work it out by themselves? How do you know the OP knows about similar triangles? – user38268 Aug 18 '11 at 3:57
@D Lim: I changed my answer. – Américo Tavares Aug 18 '11 at 8:25
@D Lim: And how do we know whether the solution is to use the formula of the area of a triangle? – Américo Tavares Aug 18 '11 at 8:29
We don't know that too. So the thing to do might be to just ask questions like what can we say about the slope of the hypotenuse? How can we measure it? Is there more than one way to measure it using the variables given above? – user38268 Aug 18 '11 at 12:58
@D Lim: There are the three measurements given by each member of $(1)$ (all equal to $h/g$). – Américo Tavares Aug 18 '11 at 13:07

Consider the area of the whole triangle and the areas of the constituent triangles and square.

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