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.
Here's a very elegant solution, based on the [proof of the angle bisector theorem](http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI3.html) that André Nicolas referred to:

Construct a line through $B$ parallel to $AD$, and extend $AC$ so that it intersects this line on $E$.

Since $AD$ is parallel to $BE$, by the [transversal theorems](http://www.dummies.com/how-to/content/definitions-and-theorems-of-parallel-lines.html) we have that both $\angle ABE$ and $\angle AEB$ are equal to $\angle BAD = 30^\circ$. Thus, $\triangle BAE$ is isosceles, and $AB = AE = 3$ (meaning $EC = 7$).

Also because $AD$ is parallel to $BE$, $\triangle CAD$ and $\triangle CEB$ are similar, meaning that $\frac {EB}{AD} = \frac {EC}{AC}$.

Drop a perpendicular from $A$ to $EB$ at $F$. Since $\triangle AEF$ is a 30-60-90 triangle, we have that $EF$ = $\frac{\sqrt{3}}{2} 3$, and $EB = 2EF = 3\sqrt{3}$. Then $\frac{3\sqrt{3}}{AD} = \frac{7}{4}$, and we have $AD$ = $\frac 47 3\sqrt{3}$, as required.