Construction of Figure Similar to One and Equal to Another

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Theorem

In the words of Euclid:

To construct one and the same figure similar to a given rectilineal figure and equal to another given rectilineal figure.

(The Elements: Book $\text{VI}$: Proposition $25$)


Construction

Let $ABC$ be the given rectilineal figure to which the figure to be constructed is to be similar, and let $D$ be the rectilineal figure to which it must be equal (in area).

Euclid-VI-25.png

Using Construction of Parallelogram on Given Line equal to Triangle in Given Angle, we apply the parallelogram $BE$ to the straight line $BC$ equal in area to $\triangle ABC$.

Using Construction of Parallelogram in Given Angle equal to Given Polygon we apply the parallelogram $CM$ to the straight line $CE$ equal in area to $D$, where $\angle FCE = \angle CBL$.

Now using Construction of Mean Proportional‎ we construct the straight line $GH$ such that:

$BC : GH = GH : CF$

Using Construction of Similar Polygon we construct $\triangle KGH$ to be similar to $\triangle ABC$ and similarly situated.

Then $\triangle KGH$ is the required rectilineal figure, similar to $\triangle ABC$ and equal in size to $D$.


Proof

We have that $BC$ is in a straight line with $CF$ and that $LE$ is in a straight line with $EM$.

From the porism to Ratio of Areas of Similar Triangles:

$BC : CF = \triangle ABC : \triangle KGH$

But from Areas of Triangles and Parallelograms Proportional to Base:

$BC : CF = \Box BE : \Box EF$

So:

$\triangle ABC : \triangle KGH = \Box BE : \Box EF$

So from Proportional Magnitudes are Proportional Alternately:

$\triangle ABC : \Box BE = \triangle KGH : \Box EF$

But:

$\triangle ABC = \Box BE$

and so:

$\triangle KGH = \Box EF$

But $\Box EF = D$, and so $\triangle KGH = D$

Also we have that $\triangle KGH$ is similar to $\triangle ABC$.

Hence the result.

$\blacksquare$


Historical Note

This proof is Proposition $25$ of Book $\text{VI}$ of Euclid's The Elements.


Sources

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