Let *ABC* be a right-angled triangle having the angle *A* right, and let the perpendicular *AD* be drawn.

I say that the rectangle *CB* by *BD* equals the square on *BA*, the rectangle *BC* by *CD* equals the square on *CA*, the rectangle *BD* by *DC* equals the square on *AD*, and the rectangle *BC* by *AD* equals the rectangle *BA* by *AC*, and first that the rectangle *CB* by *BD* equals the square on *BA*.

Since in a right-angled triangle *AD* has been drawn from the right angle perpendicular to the base, therefore the triangles *ABD* and *ADC* are similar both to the whole *ABC* and to one another.

Since the triangle *ABC* is similar to the triangle *ABD*, therefore *CB* is to *BA* as *BA* is to *BD*. Therefore the rectangle *CB* by *BD* equals the square on *AB*. For the same reason the rectangle *BC* by *CD* also equals the square on *AC*.

Since, if in a right-angled triangle a perpendicular is drawn from the right angle to the base, then the perpendicular so drawn is a mean proportional between the segments of the base, therefore *BD* is to *DA* as *AD* is to *DC*. Therefore the rectangle *BD* by *DC* equals the square on *AD*.

I say that the rectangle *BC* by *AD* also equals the rectangle *BA* by *AC*.

Since we said, *ABC* is similar to *ABD*, therefore *BC* is to *CA* as *BA* is to *AD*.

Therefore the rectangle *BC* by *AD* equals the rectangle *BA* by *AC*.

Q.E.D.

To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial.

Set out two rational straight lines *AB* and *BC* commensurable in square only such that the square on the greater *AB* is greater than the square on the less *BC* by the square on a straight line incommensurable with *AB*.

Bisect *BC* at *D*. Apply to *AB* a parallelogram equal to the square on either of the straight lines *BD* or *DC* and deficient by a square figure, and let it be the rectangle *AE* by *EB*.

Describe the semicircle *AFB* on *AB*, draw *EF* at right angles to *AB*, and join *AF* and *FB*.

Since *AB* and *BC* are unequal straight lines, and the square on *AB* is greater than the square on *BC* by the square on a straight line incommensurable with *AB*, while there was applied to *AB* a parallelogram equal to the fourth part of the square on *BC*, that is, to the square on half of it, and deficient by a square figure, making the rectangle *AE* by *EB*, therefore *AE* is incommensurable with *EB*.

And *AE* is to *EB* as the rectangle *BA* by *AE* is to the rectangle *AB* by *BE*, while the rectangle *BA* by *AE* equals the square on *AF*, and the rectangle *AB* by *BE* is to the square on *BF*, therefore the square on *AF* is incommensurable with the square on *FB*. Therefore *AF* and *FB* are incommensurable in square.

Since *AB* is rational, therefore the square on *AB* is also rational, so that the sum of the squares on *AF* and *FB* is also rational.

Since, again, the rectangle *AE* by *EB* equals the square on *EF*, and, by hypothesis, the rectangle *AE* by *EB* also equals the square on *BD*, therefore *FE* equals *BD*. Therefore *BC* is double *FE*, so that the rectangle *AB* by *BC* is also commensurable with the rectangle *ABEF*.

But the rectangle *AB* by *BC* is medial,

therefore the rectangle *AB* by *EF* is also medial.

But the rectangle *AB* by *EF* equals the rectangle *AF* by *FB*, therefore the rectangle *AF* by *FB* is also medial.

But it was also proved that the sum of the squares on these straight lines is rational.

Therefore two straight lines *AF* and *FB* incommensurable in square have been found which make the sum of the squares on them rational, but the rectangle contained by them medial.

Q.E.D.