Question: In triangle $\triangle ABC$, ray $AD$ is a bisector of angle $A$, which intersects $BC$ at $D$. Also given are that $AC$ = 4 cm, $AB$ = 3 cm and $\angle A = 60^\circ$. Find the length of $AD$.

This is a simple geometry question. I had a trivial solution using the cosine law, but my teacher said that we could only use the concept of similarity, the Pythagorean theorem, or any concepts in high school geometry excluding trigonometry. I tried for an hour but could not come up with a solution. So please give a solution with the limitations stated above. Thank you.

Answer: 
Presumably, you're allowed to know the proportions of a $30^\circ$-$60^\circ$-$90^\circ$ triangle (which arise from application of the Pythagorean Theorem to the equilateral triangle), so I'll use them.

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Let $x := |AD|$, and simply compute the area of $\triangle ABC$ in two ways:

![Two views of triangle ABC](https://i.stack.imgur.com/yuwG3.png)

$$\begin{align}
\text{area of green triangle} &= \frac{1}{2} \cdot 3\cdot 2\sqrt{3} = 3\sqrt{3} \\[6pt]
\text{area of blue triangle} &= \frac{1}{2} \cdot 3 \cdot \frac{x}{2} = \frac{3x}{4} \\[6pt]
\text{area of red triangle} &= \frac{1}{2} \cdot 4 \cdot \frac{x}{2} = x
\end{align} $$
Now,
$$\text{green} = \text{blue} + \text{red} \quad\implies\quad 3\sqrt{3} = \frac{3x}{4}+x = \frac{7x}{4} \quad\implies\quad x = \frac{12\sqrt{3}}{7}$$

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This approach certainly isn't as widely applicable as using the Law of Cosines or the Angle Bisector Theorem, but it has the advantage of being *very* direct.
The following argument is a little ugly, or perhaps a lot ugly. 

Let $AD=2x$. We will find an equation for $x$.

Drop a perpendicular from $D$ to the point $P$ on $AB$, and drop a perpendicular from $D$ to the point $Q$ on $AC$. 

Triangles $APD$ and $AQD$ are congruent. We have by basic properties of the $30^\circ$-$60^\circ$-$90^\circ$ triangle that $PD=QD=x$ and $AP=AQ=\sqrt{3}x$.

Now we use the Pythagorean Theorem for triangles $DPB$ and $CQD$ to find $BD$ and $DC$ in terms of $x$.

Note that $PB=3-\sqrt{3}{x}$, so 
$$(BD)^2=(3-\sqrt{3}x)^2+x^2\qquad\text{and}\qquad (DC)^2=(4-\sqrt{3}x)^2+x^2.\tag{1}$$

By what I think is called the **Angle Bisector Theorem,** we have $BD:DC=3:4$ and therefore $(BD)^2:(DC)^2=9:16$. 

Thus from Equations (1) we obtain
$$16\left((3-\sqrt{3}x)^2+x^2\right)=9\left((4-\sqrt{3}x)^2+x^2\right).\tag{2}.$$

Expand. The equation we get is quite simple, since constant terms cancel. Then we can cancel an $x$, and get the linear equation
$$28x=24\sqrt{3}.$$
Thus $AD=2x=\dfrac{12\sqrt{3}}{7}$.