Proof: By Euclid
(related to Proposition: 6.05: Triangles with Proportional Sides are Similar)
 For Let (angle) $FEG$, equal to angle $ABC$, and (angle) $EFG$, equal to $ACB$, have been constructed on the straight line $EF$ at the points $E$ and $F$ on it (respectively) [Prop. 1.23].
 Thus, the remaining (angle) at $A$ is equal to the remaining (angle) at $G$ [Prop. 1.32].
 Thus, triangle $ABC$ is equiangular to [triangle] $EGF$.
 Thus, for triangles $ABC$ and $EGF$, the sides about the equal angles are proportional, and (those) sides subtending equal angles correspond [Prop. 6.4].
 Thus, as $AB$ is to $BC$, [so] $GE$ (is) to $EF$.
 But, as $AB$ (is) to $BC$, so, it was assumed, (is) $DE$ to $EF$.
 Thus, as $DE$ (is) to $EF$, so $GE$ (is) to $EF$ [Prop. 5.11].
 Thus, $DE$ and $GE$ each have the same ratio to $EF$.
 Thus, $DE$ is equal to $GE$ [Prop. 5.9].
 So, for the same (reasons), $DF$ is also equal to $GF$.
 Therefore, since $DE$ is equal to $EG$, and $EF$ (is) common, the two (sides) $DE$, $EF$ are equal to the two (sides) $GE$, $EF$ (respectively).
 And base $DF$ [is] equal to base $FG$.
 Thus, angle $DEF$ is equal to angle $GEF$ [Prop. 1.8], and triangle $DEF$ (is) equal to triangle $GEF$, and the remaining angles (are) equal to the remaining angles which the equal sides subtend [Prop. 1.4].
 Thus, angle $DFE$ is also equal to $GFE$, and (angle) $EDF$ to $EGF$.
 And since (angle) $FED$ is equal to $GEF$, and (angle) $GEF$ to $ABC$, angle $ABC$ is thus also equal to $DEF$.
 So, for the same (reasons), (angle) $ACB$ is also equal to $DFE$, and, further, the (angle) at $A$ to the (angle) at $D$.
 Thus, triangle $ABC$ is equiangular to triangle $DEF$.
 Thus, if two triangles have proportional sides then the triangles will be equiangular, and will have the angles which corresponding sides subtend 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"