Construction of Figure Similar to One and Equal to Another

Theorem

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

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.