# Algebraic Manipulation in the Proof of Heron's formula

A textbook I'm reading gives a proof of Heron's formula, but has lost me in one of its steps. My mathematical foundations are a bit shaky, so I was hoping someone could explain what was done. The jump I don't understand is going from

$$\dfrac{1+\dfrac{(a^2+b^2-c^2)}{2ab}}{2}$$

to

$$\frac{a^2 + 2ab + b^2 - c^2}{4ab}$$

I tried to simplify the first equation and I ended up with

$$\frac{1}{2} + \frac{a^2 + b^2 - c^2}{4ab}$$

Is there something I've done wrong? What am I missing?

• You don't want to split out the $1$ to make $\frac12$, you want to merge it with $\frac{a^2 + b^2 + c^2}{2ab}$ to make $\frac{a^2 + 2ab + b^2 +c^2}{2ab}$. Other than that, I do think theres a typo regarding $\pm c^2$ in there, because there's no legal manipulation that will let you change that sign without changing anything else to compensate. – Arthur May 14 '15 at 0:59
• Yes, good catch - it should be $-c^2$ in both cases. – subjectification May 14 '15 at 1:01

Just notice: $$\frac12 =\frac{2ab}{4ab}$$ So both values are correct.

$$\dfrac{1+\dfrac{(a^2+b^2-c^2)}{2ab}}{2}= \dfrac{\dfrac{2ab+(a^2+b^2-c^2)}{2ab}}{2} =\dfrac{2ab+(a^2+b^2-c^2)}{4ab}.$$

Now further do as,

$$\frac{1}{2}+\frac{a^2+b^2-c^2}{4ab}$$

Taking LCM,

$$\frac{2ab}{4ab}+\frac{a^2+b^2-c^2}{4ab}$$

$$\frac{2ab+a^2+b^2-c^2}{4ab}$$