# Euclid's SSA proof (book VI prop 7)

I've been studying Euclid's Elements for the past few weeks, and have came across a conflicting theorem he proved.

"if two triangles have one angle equal to one angle, the sides about other angles proportional, and the remaining angles either both less or both not less than a right angle, the triangles will be equiangular and will have those angles equal, the sides about which are proportional."

and if you set the proportion of the corresponding sides to be 1:1, you get that the triangles are congurent! but that's wrong, as shown and told a lot to make sure that you don't show that they are congurent, while they're not.

so how did Euclid prove this theorem, if it's wrong? What's the error in Euclid's proof?

• In your example, $\widehat{CAB}>\pi/2$ and $\widehat{CBA}<\pi/2$? Commented Jun 23 at 10:12
• both less and not less then a right angle, meaning $\widehat{CAB}\lt\frac{\pi}{2}$ and $\widehat{CAB}\ge\frac{\pi}{2}$ Commented Jun 23 at 10:19

Reading through the proof you gave a link to, it seems that the statement must be understood as: let $$\triangle ABC$$ and $$\triangle DEF$$ be two triangle satisfying:
1. $$\widehat{ACB}=\widehat{EFD}$$ ;
2. $$|BC|/|AB|=|EF|/|DE|$$ ;
3. either $$\widehat{BAC}>\pi/2$$ and $$\widehat{EDF}>\pi/2$$ or $$\widehat{BAC}<\pi/2$$ and $$\widehat{EDF}<\pi/2$$.
Then $$\triangle ABC$$ and $$\triangle DEF$$ are equiangular.
• The first condition should be $\angle ACB =\angle EFD$. Commented Jun 23 at 16:26