If in a right-angled triangle a perpendicular is drawn from the right angle to the base, then the triangles adjoining the perpendicular are similar both to the whole and to one another.

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

I say that each of the triangles *DBA* and *DAC* is similar to the whole *ABC,* and, further, they are similar to one another.

Since the angle *BAC* equals the angle *BDA,* for each is right, and the angle at *B* is common to the two triangles *ABC* and *DBA,* therefore the remaining angle *ACB* equals the remaining angle *DAB.* Therefore the triangle *ABC* is equiangular with the triangle *DBA.*

Therefore *BC,* which is opposite the right angle in the triangle *ABC,* is to *BA,* which is opposite the right angle in the triangle *DBA,* as *AB,* which is opposite the angle at *C* in the triangle *ABC,* is to *DB,* which is opposite the equal angle *BAD* in the triangle *DBA,* and also as *AC* is to *DA,* which is opposite the angle at *B* common to the two triangles.

Therefore the triangle *ABC* is both equiangular to the triangle *DBA* and has the sides about the equal angles proportional.

Therefore the triangle *ABC* is similar to the triangle *DBA.*

In the same manner we can prove that the triangle *DAC* is also similar to
the triangle *ABC.* Therefore each of the triangles *DBA* and *DAC* is similar to the whole *ABC.*

I say next that the triangles *DBA* and *DAC* are also similar to one another.

Since the right angle *BDA* equals the right angle *ADC,* and moreover the angle *DAB* was also proved equal to the angle at *C,* therefore the remaining angle at *B* also equals the remaining angle *DAC.* Therefore the triangle *DBA* is equiangular with the triangle *ADC.*

Therefore *BD,* which is opposite the angle *DAB* in the triangle *DBA,* is to *AD,* which is opposite the angle at *C* in the triangle *DAC* equal to the angle *DAB,* as *AD,* itself which is opposite the angle at *B* in the triangle *DBA,* is to *CD,* which is opposite the angle *DAC* in the triangle *DAC* equal to the angle at *B,* and also as *BA* is to *AC,* these sides opposite the right angles. Therefore the triangle *DBA* is similar to the triangle *DAC.*

Therefore, *if in a right-angled triangle a perpendicular is drawn from the right angle to the base, then the triangles adjoining the perpendicular are similar both to the whole and to one another.*

Q.E.D.

From this it is clear that, if in a right-angled triangle a perpendicular is drawn from the right angle to the base, then the straight line so drawn is a mean proportional between the segments of the base.

Note that Euclid verbosely draws from proposition VI.4 the conclusions that equiangular triangles are similar and that triangles similar to the same triangle are similar to each other. The general proposition that figures similar to the same figure are also similar to one another is proposition VI.21. There is no reason why that proposition could not have been placed before this one.

This proposition may be used to give an alternate proof of proposition I.47. Indeed, Euclid presents such a proof in the lemma for X.33. That proof is probably older than Euclid’s as given in I.47, but Euclid’s proof has the advantage of not being dependent on Eudoxus’ theory of proportion in Book V.

This proposition and its corollary are used in propositions VI.13, VI.31, X.33, and often in Book XIII.