Proof: By Euclid
(related to Lemma: Lem. 10.032: Constructing Medial Commensurable in Square II)
- For since AD has been drawn from the right angle in a right-angled triangle, perpendicular to the base, ABD and ADC are thus triangles (which are) similar to the whole, ABC, and to one another [Prop. 6.8].
- And since triangle ABC is similar to triangle ABD, thus as CB is to BA, so BA (is) to BD [Prop. 6.4].
- Thus, the (rectangle contained) by CBD is equal to the (square) on AB [Prop. 6.17].
- So, for the same (reasons), the (rectangle contained) by BCD is also equal to the (square) on AC.
- And since if a (straight line) is drawn from the right angle in a right-angled triangle, perpendicular to the base, the (straight line so) drawn is in mean proportion3 to the pieces of the base [Prop. 6.8 corr.] 3, thus as BD is to DA, so AD (is) to DC.
- Thus, the (rectangle contained) by BD and DC is equal to the (square) on DA [Prop. 6.17].
- I also say that the (rectangle contained) by BC and AD is equal to the (rectangle contained) by BA and AC.
- For since, as we said, ABC is similar to ABD, thus as BC is to CA, so BA (is) to AD [Prop. 6.4].
- Thus, the (rectangle contained) by BC and AD is equal to the (rectangle contained) by BA and AC [Prop. 6.16].
- (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"