---
book: 6
number: 8
id: "VI.8"
kind: "theorem"
uses: ["[[book-1/proposition-32]]", "[[book-6/proposition-4]]"]
source: "https://scaife.perseus.org/reader/urn:cts:greekLit:tlg1799.tlg001.perseus-eng2:6.prop.8"
license: "CC-BY-SA-4.0"
---

# VI.8

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

## Proof

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 *ABD*, *ADC* is similar to the whole *ABC* and, further, they are similar to one another.

For, since the angle *BAC* is equal to the angle *ADB*, for each is right, and the angle at *B* is common to the two triangles *ABC* and *ABD*, therefore the remaining angle *ACB* is equal to the remaining angle *BAD*; [[book-1/proposition-32|I. 32]] therefore the triangle *ABC* is equiangular with the triangle *ABD*.

Therefore, as *BC* which subtends the right angle in the triangle *ABC* is to *BA* which subtends the right angle in the triangle *ABD*, so is *AB* itself which subtends the angle at *C* in the triangle *ABC* to *BD* which subtends the equal angle *BAD* in the triangle *ABD*, and so also is *AC* to *AD* which subtends the angle at *B* common to the two triangles. [[book-6/proposition-4|VI. 4]]

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

Therefore the triangle *ABC* is similar to the triangle *ABD*. [[book-6/definitions#Definition 1|VI. Def. 1]]

Similarly we can prove that the triangle *ABC* is also similar to the triangle *ADC*; therefore each of the triangles *ABD*, *ADC* is similar to the whole *ABC*.

I say next that the triangles *ABD*, *ADC* are also similar to one another.

For, since the right angle *BDA* is equal to the right angle *ADC*, and moreover the angle *BAD* was also proved equal to the angle at *C*, therefore the remaining angle at *B* is also equal to the remaining angle *DAC*; [[book-1/proposition-32|I. 32]] therefore the triangle *ABD* is equiangular with the triangle *ADC*.

Therefore, as *BD* which subtends the angle *BAD* in the triangle *ABD* is to *DA* which subtends the angle at *C* in the triangle *ADC* equal to the angle *BAD*, so is *AD* itself which subtends the angle at *B* in the triangle *ABD* to *DC* which subtends the angle *DAC* in the triangle *ADC* equal to the angle at *B*, and so also is *BA* to *AC*, these sides subtending the right angles; [[book-6/proposition-4|VI. 4]] therefore the triangle *ABD* is similar to the triangle *ADC*. [[book-6/definitions#Definition 1|VI. Def. 1]]

Therefore etc.

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