Simple geometry question, to be proved without trigonometry

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 1

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:

$$\begin{align} \text{area of green triangle} &= \frac{1}{2} \cdot 3\cdot 2\sqrt{3} = 3\sqrt{3} \\[6pt] \text{area of blue triangle} &= \frac{1}{2} \cdot 3 \cdot \frac{x}{2} = \frac{3x}{4} \\[6pt] \text{area of red triangle} &= \frac{1}{2} \cdot 4 \cdot \frac{x}{2} = x \end{align} $$ 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.

Answer 2

(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 (which is proved here 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).

Answer 3

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 $AD=2x=\dfrac{12\sqrt{3}}{7}$.

Answer 4

Here's a very elegant solution, based on the proof of the angle bisector theorem 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 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.