Triangle Angle-Side-Angle and Side-Angle-Angle Equality

Theorem

Part 1

If two triangles have

• Two angles equal to two angles, respectively;
• The sides between the two angles equal

Then the remaining angles are equal, and the remaining sides equal the respective sides.
That is to say, if two pairs of angles and the included sides are equal, then the triangles are equal.

Part 2

If two triangles have

• Two angles equal to two angles, respectively;
• The sides opposite one pair of equal angles equal

Then the remaining angles are equal, and the remaining sides equal the respective sides.
That is to say, if two pairs of angles and a pair of opposite sides are equal, then the triangles are equal.

Proof

Part 1

Let ∠ABC=∠DEF , ∠BCA=∠EFD , and BC=EF .
Assume AB≠DE . If this is the case, one of the two must be greater. WLOG, we let AB>DE .
We construct a point G on AB such that BG=ED , and then we construct the segment CG .
Now, since we have BG=ED , ∠GBC=∠DEF , and BC=EF , from Triangle Side-Angle-Side Equality we have ∠GCB=∠DFE .
But from Euclid's fifth common notion ∠DFE=∠ACB>∠GCB , a contradiction.
Therefore, AB=DE , so from Triangle Side-Angle-Side Equality, we have △ABC=△DEF .
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Part 2

Let ∠ABC=∠DEF , ∠BCA=∠EFD , and AB=DE .
Assume BC≠EF . If this is the case, one of the two must be greater. WLOG, we let BC>EF .
We construct a point H on BC such that BH=EF , and then we construct the segment AH .
Now, since we have BH=EF , ∠ABH=∠DEF , and AB=DE , from Triangle Side-Angle-Side Equality we have ∠BHA=∠EFD .
But from External Angle of Triangle Greater than Internal Opposite, we have ∠BHA>∠HCA=∠EFD , a contradiction.
Therefore, BC=EF , so from Triangle Side-Angle-Side Equality, we have △ABC=△DEF .
■

Alternate Proof

Either part of this theorem follows trivially from the other part and the fact that the sum of the angles of a triangle equals two right angles. However, it is important to note that both of these are provable without the parallel postulate, which the proof of that theorem requires.
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