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.
(Note: I've since posted a better answer, but I'm keeping this one here for historical purposes, since one of the comments on this answer led me to create the other one.)

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Choose a point $E$ on $AC$ so that $AE$ = 3 cm. Then, note that $ABE$ is equilateral, and $AD$, the angle bisector of $\angle A$, is a perpendicular bisector of $BE$, intersecting it at $F$. Therefore, $\triangle ABF$ is a right triangle.

Since $\triangle ABE$ is equilateral, and $BF$ = $\frac 12 AB$, by the Pythagorean theorem we have $AF^2 + BF^2 = AB^2$, which means that $AF = \frac{3\sqrt{3}}{2}$.

Similarly, if we drop a perpendicular from $B$ to $AE$, we see that its length is also $\frac{3\sqrt{3}}{2}$, so the area of the triangle $ABC$ is $\frac 12 \ 4\ \frac {3\sqrt{3}}{2} = 3\sqrt{3}$.

Now, by something called the [angle bisector theorem](http://en.wikipedia.org/wiki/Angle_bisector_theorem) (which is proved [here](http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI3.html) without using similar triangles and not trigonometry), we can deduce that $\triangle ADC$ has $\frac 47$ the area of the big triangle, and by noting that $AE = 3$ and $EC = 1$, we see that $\triangle DEC$ has $\frac 14$ the area of that triangle, so $\triangle DEC$ has an area of $\frac 17 3\sqrt{3}$.

So the quadrilateral $ABDE$ has an area of $\frac 67 3\sqrt{3}$, and we note that since $AD$ is a perpendicular bisector of $BE$, that $ABDE$ is actually a kite. Then, the length of $AD$ is just the area of $ABDE$ divided by $BE$ and divided again by $\frac 12$, which works out to be $\frac 47 3\sqrt{3}$.

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This solution makes use of only the Pythagorean Theorem, similar triangles, and other high school geometry concepts, as required by your teacher.

Thanks to André Nicolas for finding the non-trigonometric proof of the angle bicsector theorem (which I honestly think is much more elegant than the trigonometric proof).