# showing subtraction of angles from addition of angles axiom in geometry

Suppose we are given congruent angles $\angle BAC \cong\angle B'A'C'$

Suppose also that we are given a ray AD in the interior of $\angle BAC$

Show that then there exists a ray A'D' in the interior of $\angle B'A'C'$ such that $\angle DAC \cong \angle D'A'C'$ and $\angle BAD \cong \angle B'A'D'$

This statement corresponds to Euclid's Common Notion 3: "Equals subtracted from equals are equal," where "equal" in this case means congruence of angles.

This is supposed to be an easy exercise in Hartshorne (9.1) but i cant seem to figure it out.

## 1 Answer

I'll base my development on this pdf file. Just a warning, but I was way too lazy to check if everything I say fits in with the definitions and axioms used to define angles, so this is at most something to get you started? If you can point me to the place in the book/pdf where they specify what it means for a ray to be inside an angle, I could try to make this better I guess.

[Rant] My answer seems way too complicated, but I really don't feel like getting into the content of the book to look for shortcuts.

According to axiom $(C4)$, there exists two points $D'_1$ and $D'_2$ such that:

• the rays $A'D'_1$ and $A'D'_2$ are on either side of the line $A'B'$, and
• $\angle B'A'D'_i\cong \angle BAD$.

I argue that one of those two points satisfy the desired property.

To see this, we make use of axiom $(C4)$ again. Take one of the point $D'_i$, there must exist a point $C''_i$ such that $\angle D'_iA'C''_i \cong \angle DAC$. We have the choice between two such point $C''_i$ but we choose $C''_i$ so that it is on one side of the ray $A'D'_i$, and point $B'$ is on the other side of that ray. We can now apply Proposition 9.4 (Addition of angles): the rays $A'B'$ and $A'C''_i$ form an angle, and $\angle B'A'C''_i \cong \angle BAC$. Then, axiom $(C5)$, or Proposition 9.1 as well, yields that $\angle B'A'C''_i \cong \angle B'A'C'$.

I claim that the two points $C''_1$ and $C''_2$ define two rays $A'C_1''$ and $A'C_2''$ that lie on opposite sides of line $A'B'$. Suppose by contradiction that they do not, axiom $(C4)$ exactly states that:

Given an angle $\angle BAC$ and given a ray $DF$, there exists a unique ray $DE$, on a given side of the line $DF$, such that $\angle BAC \cong \angle EDF$.

It follows that the rays $A'C''_1$ and $A'C''_2$ are equal. Because we built $C''_i$ using Proposition 9.4 (Addition of angles), we know that the ray $A'D'_i$ is in the interior of angle $\angle B'A'C''_i$. I assume that from there, we can deduce that rays $A'D'_1$ and $A'D'_2$ are on the same side of line $A'B'$, contradicting the definition of $D'_1$ and $D'_2$. This argument is rather imprecise because I didn't check what "a ray being inside an angle" precisely mean. But I have a hunch you could make this more formal using congruent triangles. In fact we can force the two triangles $A'B'C''_1$ and $A'B'C_2''$ to be equal, then force point $D'_i$ to lie on the line $B'C_i''$. From there, you can probably show that $D'_1=D'_2$ using some results on the segments that come earlier in the book.

Anyway, let's say that we prove one way or another that rays $A'C_1''$ and $A'C_2''$ are indeed on opposite sides of line $A'B'$. If you give me that, then we can move forward with axiom $(C4)$.

Because $\angle B'A'C_i'' \cong \angle B'A'C'$, and the two rays $A'C_1''$ and $A'C_2''$ are on opposite sides of line $A'B'$, one of these two rays must coincide with ray $A'C'$. This gives $\angle D'_iA'C''_i \cong \angle D'_iA'C'$. This means that point $D'_i$ satisfy the two angle congruences you want:

• $\angle BAD \cong \angle B'A'D'_i$
• $\angle DAC \cong \angle D_i'A'C'$

Remains to prove that ray $A'D'_i$ is in the interior of $\angle B'A'C'$. Again, I'm not too sure here.

Intuitively, I'd say this is feasible if you do some hackery with congruent triangles (again). Make $A'B'C'$ congruent to $ABC$. Force $D$ to be on line $BC$. When Building $D'_i$, make $D'_iA'B'$ congruent to $DAB$. You can probably prove that $D'_i$ lies inside the line segment between $C'$ and $B'$. I'm not sure if that is enough, but well...