# Prove Right Angle Side Side (RASS) congruence (from Hartshorne)

Prove if two right triangles have one side and the hypotenuse respectively congruent, then the triangles are congruent (from Hartshorne, Geometry: Euclid and Beyond).

Proof: Let the triangles be $$\triangle ABC$$ and $$\triangle A'B'C'$$, with angles $$\angle B, \angle B'$$ right angles, hypotenuses $$AC \cong A'C'$$, and sides $$AB \cong A'B'$$.

Extend $$AB$$ to $$D$$ such that $$AB \cong BD$$. Angle $$\angle CBD$$ is also a right angle, and so $$\triangle ABC$$ and $$\triangle DBC$$ are congruent by SAS, and $$DC \cong AC$$.

Similarly extend $$A'B'$$ to $$D'$$, which results in $$\triangle ADC \cong \triangle A'D'C'$$ by SSS.

Observe that $$\angle ACD \cong \angle A'C'D'$$, with each angle bisected by $$BC$$ or $$B'C'$$ respectively. Consequently, $$\angle ACB \cong \angle A'C'B'$$, and $$\triangle ACB \cong A'C'B'$$ by AAS.

Questions: Is the proof correct and well-written? Is the technique of reflecting the triangle over the unknown leg a good one? Is there a simpler approach? Please verify, critique, or improve the proof and its exposition.

## Update

Based on Blue's suggestion, here is a simpler conclusion to the proof:

Similarly extend $$A'B'$$ to $$D'$$, which results in $$\triangle ADC \cong \triangle A'D'C'$$ by SSS. Thus, $$\angle A \cong \angle A'$$ and, by SAS, $$\triangle ABC \cong \triangle A'B'C'$$.

• I believe if you're going to use SSS,SAS AAS in the proof, you may as well just prove it with SAS, since you know that <B is congruent with <B' (Angle), AB is congruent with A'B' (Side #1) and AC is congruent with A'C' (Side #2). Perhaps that works too? Commented Jan 26, 2023 at 2:58
• @MathMan I don't believe that works: Since $\angle B$ is not between $AB$ and $AC$, that would be SSA, which is not always congruent. To use SAS this way, we'd need to know $\angle A \cong \angle A'$. Commented Jan 26, 2023 at 3:01
• @SRobertJames: Your proof looks okay to me, provided the notion that halves of congruent angles are congruent is in play. (At the foundations level, it can be tricky to know what's allowed, since different authors can present results in different orders.) Note that your earlier comment to MathMan suggests a way forward that's "safer" in this regard: once you have $\triangle ADC\cong\triangle A'D'C'$, you can say $\angle A\cong\angle A'$ so that $\triangle ABC\cong\triangle A'B'C'$ by SAS. No subtle angle-algebra required.
– Blue
Commented Jan 26, 2023 at 3:42
• @Blue Your suggestion also makes the proof simpler and shorter - all win, no downside. Revised accordingly. (PS If you make your comment an answer, I will accept it.) Commented Jan 26, 2023 at 3:56

Note that your comment to MathMan suggests a way forward that's "safer" in this regard: Once you have $$\triangle ADC\cong\triangle A'D'C'$$, you can say $$\angle A\cong\angle A'$$ so that $$\triangle ABC\cong\triangle A'B'C'$$ by SAS. No subtle angle-algebra required.