To construct a figure similar to one given rectilinear figure and equal to another.

Let *ABC* be the given rectilinear figure to which the figure to be constructed must be similar, and *D* that to which it must be equal.

It is required to construct one figure similar to *ABC* and equal to *D.*

Let there be applied to *BC* the parallelogram *BE* equal to the triangle *ABC,* and to *CE* the parallelogram *CM* equal to *D* in the angle *FCE* which equals the angle *CBL.*

Then *BC* is in a straight line with *CF,* and *LE* with *EM.*

Take a mean proportional *GH* to *BC* and *CF,* and describe *KGH* similar and similarly situated to *ABC* on *GH.*

Then, since *BC* is to *GH* as *GH* is to *CF,* and, if three straight lines are proportional, then the first is to the third as the figure on the first is to the similar and similarly situated figure described on the second, therefore *BC* is to *CF* as the triangle *ABC* is to the triangle *KGH.*

But *BC* is to *CF* as the parallelogram *BE* is to the parallelogram *EF.*

Therefore also the triangle *ABC* is to the triangle *KGH* as the parallelogram *BE* is to the parallelogram *EF.* Therefore, alternately, the triangle *ABC* is to the parallelogram *BE* as the triangle *KGH* is to the parallelogram *EF.*

But the triangle *ABC* equals the parallelogram *BE,* therefore the triangle *KGH* also equals the parallelogram *EF.* And the parallelogram *EF* equals *D,* therefore *KGH* also equals *D.*

And *KGH* is also similar to *ABC.* Therefore this figure *KGH* has been constructed similar to the given rectilinear figure *ABC* and equal to the other given figure *D.*

Q.E.F.

This proposition solves a similar problem, to find a figure with the size of one figure but the shape of another, a problem reputedly solved by Pythagoras. It is used in the proofs of propositions VI.28 and VI.29