Proof: By Euclid
(related to Proposition: 6.15: Characterization of Congruent Triangles)
- For let $CA$ be laid down so as to be straight-on (with respect) to $AD$.
- Thus, $EA$ is also straight-on (with respect) to $AB$ [Prop. 1.14].
- And let $BD$ have been joined.
- Therefore, since triangle $ABC$ is equal to triangle $ADE$, and $BAD$ (is) some other (triangle), thus as triangle $CAB$ is to triangle $BAD$, so triangle $EAD$ (is) to triangle $BAD$ [Prop. 5.7].
- But, as (triangle) $CAB$ (is) to $BAD$, so $CA$ (is) to $AD$, and as (triangle) $EAD$ (is) to $BAD$, so $EA$ (is) to $AB$ [Prop. 6.1].
- And thus, as $CA$ (is) to $AD$, so $EA$ (is) to $AB$.
- Thus, in triangles $ABC$ and $ADE$ the sides about the equal angles (are) reciprocally proportional.
- And so, let the sides of triangles $ABC$ and $ADE$ be reciprocally proportional, and (thus) let $CA$ be to $AD$, as $EA$ (is) to $AB$.
- I say that triangle $ABC$ is equal to triangle $ADE$.
- For, $BD$ again being joined, since as $CA$ is to $AD$, so $EA$ (is) to $AB$, but as $CA$ (is) to $AD$, so triangle $ABC$ (is) to triangle $BAD$, and as $EA$ (is) to $AB$, so triangle $EAD$ (is) to triangle $BAD$ [Prop. 6.1], thus as triangle $ABC$ (is) to triangle $BAD$, so triangle $EAD$ (is) to triangle $BAD$.
- Thus, (triangles) $ABC$ and $EAD$ each have the same ratio to $BAD$.
- Thus, [triangle] $ABC$ is equal to triangle $EAD$ [Prop. 5.9].
- Thus, in equal triangles also having one angle equal to one (angle) the sides about the equal angles (are) reciprocally proportional.
- And those triangles having one angle equal to one angle for which the sides about the equal angles (are) reciprocally proportional are equal.
- (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"
