Given $\triangle ABC$ with $\angle C = 60^\circ$, show $a^2+b^2-ab=c^2$ without trigonometry

Question

I have the following problem:

In $\triangle ABC$, the angle at vertex $C$ is $60^{\circ}$. Prove that $a^2+b^2-ab=c^2$.

Of course, it is easy if you use cosine rule. I believe there exists a beautiful proof as well without using trigonometry. Anyone?

Answer 1

Hint

Using the picture you get $q=p$.

Now think about the next picture:

Answer 2

Drop perpendicular from $C$ onto $AB$ and call foot of perpendicular as $D$. Then using pythagoras theorem, $$AD^2 + DC^2 = b^2 \tag{1}$$ $$DC^2 + DB^2 = a^2\tag{2}$$ and from side $c$ $$AD + DB = c \tag{3}$$

Finally we need a relation that comes from the angle $A$ being $60^\circ$. Consider the triangle $ADC.$ Make a midpoint $E$ on $AC$ and join $ED$. Thus from what Arnaldo showed, we can show $$AD = \frac{b}{2}\tag{4}$$

So you have to just eliminate between these equations.

Answer 3

Without using trigonometry, we can use Stewart's theorem.

We build an equilateral triangle as shown in the attached figure so we have

$$b^2(a-b)+c^2b=ab(a-b)+b^2a$$ which is equivalent to the given $$a^2+b^2-ab=c^2$$

Answer 4

Let $a\leq b$ and $BD$ be an altitude of the triangle.

Thus, since $$\measuredangle A\leq\frac{180^{\circ}-60^{\circ}}{2}=60^{\circ},$$ we see that $D$ is placed between $A$ and $C$.

Hence, $DC=\frac{a}{2}$, $AD=b-\frac{a}{2}$, $BD=\frac{\sqrt3a}{2}$ and by Pythagoras we obtain: $$c^2=\left(b-\frac{a}{2}\right)^2+\left(\frac{\sqrt3a}{2}\right)^2=a^2-ab+b^2.$$

Answer 5

Here's a proof using Ptolemy's Theorem:

$$\begin{align} |\overline{AB}| |\overline{A^\prime B^\prime}| &= |\overline{AA^\prime}||\overline{BB^\prime}|+|\overline{AB^\prime}||\overline{A^\prime B}| \tag{1} \\[4pt] c^2 &= a b + ( a - b )^2 \tag{2} \\[4pt] &= a^2 + b^2 - a b \tag{3} \end{align}$$

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Incidentally, the Wikipedia entry mentions that one can use Ptolemy to prove the Law of Cosines generally. However, the description there isn't as clear as it might be. All one needs to do is observe that, in general, $$|\overline{AA^\prime}| = 2b\sin\frac{C}{2} \qquad |\overline{BB^\prime}| = 2a\sin\frac{C}{2} \tag{4}$$ so that the counterpart of $(2)$ is $$c^2 = 4ab\sin^2\frac{C}{2} + ( a - b )^2 \tag{5}$$ and we have $$c^2 = a^2 + b^2 - 2 a b \left( 1 - 2\sin^2\frac{C}{2} \right) = a^2 + b^2 - 2 a b \cos C \tag{6}$$