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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.

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    $\begingroup$ Please don't use all caps in your title or post. It's considered shouting, and rude. Regardless, please share your thoughts on the problem and what you've tried. $\endgroup$ – user61527 Sep 8 '13 at 4:06
  • $\begingroup$ my english is weak. that's why i am typing this type. I do not want to shouting and not talking rudely with you. Please, sorry. I will consider your instructions next time. $\endgroup$ – Hardey Pandya Sep 8 '13 at 7:41
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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.


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

Two views of triangle ABC

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


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.

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  • $\begingroup$ The blue-red figure suggests a non-trigonometric proof of the Angle Bisector Theorem. Regardless of the angle, the altitudes from $D$ to $AB$ and $BC$ in the blue and red triangles necessarily match (as corresponding parts of congruent right triangles); writing $h$ for the length, we have $$\text{blue area} = h\;|AB|/2 \qquad \text{red area} = h\;|AC|/2$$But also, writing $k$ for the length of the altitude dropped from $A$ to $BC$, we have $$\text{blue area} = k\;|BD|/2 \qquad \text{red area} = k\;|CD|/2$$ Consequently, $$\frac{|AB|}{|AC|}=\frac{\text{blue}}{\text{red}}=\frac{|BD|}{|CD|}$$ $\endgroup$ – Blue Sep 8 '13 at 6:21
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    $\begingroup$ I think this one wins the contest of using the least machinery. $\endgroup$ – Joe Z. Sep 8 '13 at 10:02
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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.

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  • $\begingroup$ Nice one!${}{}{}{}{}$ $\endgroup$ – André Nicolas Sep 8 '13 at 5:14
  • $\begingroup$ Thanks. If it weren't for you, I wouldn't have seen the resource necessary to come up with it. $\endgroup$ – Joe Z. Sep 8 '13 at 5:16
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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 $AD=2x=\dfrac{12\sqrt{3}}{7}$.

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  • $\begingroup$ Your solution is quite algebra-heavy, I think. $\endgroup$ – Joe Z. Sep 8 '13 at 4:51
  • $\begingroup$ Ihe last equation is very much on the heavy handed side! $\endgroup$ – André Nicolas Sep 8 '13 at 5:07
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(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.)


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 $\frac 47 3\sqrt{3}$.


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).

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  • $\begingroup$ I have a feeling a more elegant solution is available; however, I couldn't come up with it. $\endgroup$ – Joe Z. Sep 8 '13 at 4:31
  • $\begingroup$ Although the "easy" way to the angle bisector theorem is via the Sine Law, there are purely synthetic proofs. Have seen it, have even done it in the dim past. But am lousy at searching. $\endgroup$ – André Nicolas Sep 8 '13 at 4:37
  • $\begingroup$ I see. A quick google for "angle bisector theorem non-trigonometric proof" returned nothing useful, unfortunately. $\endgroup$ – Joe Z. Sep 8 '13 at 4:38
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    $\begingroup$ Euclid's Elements, Book VI, Prop 3! $\endgroup$ – André Nicolas Sep 8 '13 at 4:41

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