Prove side $c$ of a Triangle equals $\frac{0}{2}$

Question

So in $\triangle ABC$ with angle $\theta$ we can calculate $$c^2 = a^2 + b^2 - 2ab\cos \theta $$ $$\cos\theta = \frac{a^2 + b^2 - c^2}{2ab} $$ $$\Longrightarrow$$ $$c^2 = a^2 + b^2 - 2ab\frac{a^2 + b^2 - c^2}{2ab}$$ $$2ab\frac{a^2 + b^2 - c^2}{2ab} = a^2 + b^2 - c^2$$ $$\Longrightarrow$$ $$c^2 = a^2 + b^2 - a^2 + b^2 - c^2$$ $$c^2+c^2 = a^2 + b^2 - a^2 + b^2$$ $$2c^2 = a^2 + b^2 - a^2 + b^2$$ $$c = \sqrt{\frac{0}{2}}$$ How did this happen? I'm sure I did something wrong.

EDIT:
I just forgot the brackets $$c^2 = a^2 + b^2 - (a^2 + b^2 - c^2)$$

Answer 1

BRACKETS!!!

$$c^2 = a^2 + b^2 - 2ab\frac{a^2 + b^2 - c^2}{2ab}$$ $$c^2 = a^2 + b^2 - (a^2 + b^2 - c^2)$$ $$c^2 = a^2 + b^2 - a^2 - b^2 + c^2$$ $$c^2 = c^2$$

This is an excellent question to learn to use brackets :-)

Answer 2

$$2ab\frac{a^2 + b^2 - c^2}{2ab} = a^2 + b^2 - c^2$$ $$\Longrightarrow$$ $$c^2 = a^2 + b^2 - a^2 + b^2 - c^2$$

These two equations are not equivalent at all. You messed up not one, but two signs.

The original equation simplifies to $$a^2+b^2-c^2 = a^2+b^2-c^2$$ which is trivially true, but also does not imply $c=0$.

---

Also, as advice, if you get similar nonsensical equations in the future, I suggest you plug in numbers from an actual triangle. For example, take a very simple case where $a=b=c=1$, in which case you also know that $\phi=\frac\pi3=60^\circ$ and $\cos\phi = \frac12$. Now check which equations are still valid, and which are not:

$$\begin{align} c^2 &= a^2 + b^2 - 2ab\cos\theta & 1^2&=1^2+1^2-2\cdot 1\cdot 1\cdot 1\cdot\frac12 \\ \cos\theta &= \frac{a^2 + b^2 - c^2}{2ab} & \frac12 &= \frac{1^2+1^2-1^2}{2\cdot 1\cdot 1} \\ \Longrightarrow\\ c^2 &= a^2 + b^2 - 2ab\frac{a^2 + b^2 - c^2}{2ab} & 1^2&=1^2+1^2-2\cdot1\cdot1\cdot\frac{1^2+1^2-1^2}{2\cdot 1 \cdot 1} \\ 2ab\frac{a^2 + b^2 - c^2}{2ab} &= a^2 + b^2 - c^2 & 2 \cdot 1 \cdot 1 \cdot \frac{1^2+1^2-1^2}{2 \cdot 1 \cdot 1} &= 1^2+1^2-1^2 \\ \Longrightarrow \\ c^2 &= a^2 + b^2 - a^2 + b^2 - c^2 & 1^2&=1^2+1^2-1^2+1^2-c^2 \\ c^2+c^2 &= a^2 + b^2 - a^2 + b^2 & 1^2+1^2&=1^2+1^2-1^2+1^2 \\ 2c^2 &= a^2 + b^2 - a^2 + b^2 & 2 \cdot 1^2&=1^2+1^2-1^2+1^2 \\ c &= \sqrt{\frac{0}{2}}&&1=\sqrt{\frac02}\\ \end{align}$$