# Proof: By Euclid

• 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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