G.24 congruent triangles by SAS — Heath¶
Let < A B C >, < D E F > be two triangles having the two sides / A B /, / A C / equal to the two sides / D E /, / D F / respectively, namely / A B / to / D E / and / A C / to / D F /, and the angle < B A C ) equal to the angle < E D F ).
I say that the base / B C / is also equal to the base / E F /, the triangle < A B C > will be equal to the triangle < D E F >, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend, that is, the angle < A B C ) to the angle < D E F ), and the angle < A C B ) to the angle < D F E ).
For, if the triangle < A B C > be applied to the triangle < D E F >, and if the point A be placed on the point D and the straight line / A B / on / D E /, then the point B will also coincide with E, because / A B / is equal to / D E /.
Again, / A B / coinciding with / D E /, the straight line / A C / will also coincide with / D F /, because the angle < B A C ) is equal to the angle < E D F ); hence the point C will also coincide with the point F, because / A C / is again equal to / D F /.
But B also coincided with E; hence the base / B C / will coincide with the base / E F / [For if, when B coincides with E and C with F, the base / B C / does not coincide with the base / E F /, two straight lines will enclose a space: which is impossible. Therefore the base / B C / will coincide with / E F /] and will be equal to it. [G.23]
Thus the whole triangle < A B C > will coincide with the whole triangle < D E F >,
And the remaining angles will also coincide with the remaining angles and will be equal to them,
Therefore etc.
Dependency Graph¶
Required for¶
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