# In $\triangle ABC$, $\angle A=60^\circ$ and $\angle B=30^\circ$ and $AD$ the bisector of $\angle A=60^\circ$. Find $\frac{AD}{BC}=?$

In the triangle $$\triangle ABC$$ we have: $$\angle A=60^\circ$$ $$\angle B=30^\circ$$ And bisector of the angle $$\angle A=60^\circ$$ is $$AD$$ that divides it into 2 equal parts: $$\angle CAD=30^\circ$$ $$\angle DAB=30^\circ$$ So we want now find: $$\frac{AD}{BC}=?$$ I have written all trigonometric relations between the sides. But unhopefully no results. Would you hint me about these relations?

• Yeah but everyone knows them! Oct 27, 2018 at 19:16
• Have you included all the trigonometric relations between the sides in you post? I don't see those. I see info about angles, only. Please do not leave out work you've done that might prove crucial to solving the question. We know, by deduction, that $\triangle ABC$ is a right triangle, given $|\angle A| + |\angle B| = 90^\circ$. That is, $\angle C = 180^\circ - 90^\circ = 90^\circ$. Please include more of the work you did, as an edit to your post. Oct 27, 2018 at 19:19
• At first you must compute the length of $$AD$$ by the Pythagorean theorem Oct 27, 2018 at 19:23
• So can you have a complete answer!? Oct 27, 2018 at 19:26
• Sorry, but my time is limited too. I don't necessarily cater to users who ask urgent questions for homework due the next day, or have an exam on Sunday?. Please don't ask more of answerers than you're willing to do in a question. Please read the helpful thread: How to ask a good question on math.se, (if you want a good answer). I'll have to go know. Oct 27, 2018 at 19:37

Since $$AC:AB=1:2$$ we know that $$D$$ as bisector divides $$CB$$in the same relation, so $$D$$ divides $$CB$$ in relation $$1:2$$. From $$\angle(DAC)=\angle(ABD)=30^{\circ}$$ we conclude $$AD=BD$$ and finally we have we have $$AD:BC=BD:BC=2:3$$.

Yet another solution: reflect the triangle on $$BC$$ to get an equilateral triangle $$ABA’$$; the prolongation of $$AD$$ meets $$A’B$$ in $$D’$$. Now $$AD’$$ and $$BC$$ are also medians with the same length and their intersection $$D$$ divides each in relation $$1:2$$ again.

• Typographical error: You meant to write $D$ as bisector divides $CB$ in the same relation. Oct 28, 2018 at 2:45
• Thanks, corrected. Oct 28, 2018 at 9:42

It is good that you wrote all trigonometric relations between the sides. Now you should relate them to each other. There are so many relations one can find.

Refer to the figure:

$$\hspace{5cm}$$

Method $$1$$: (Michael Hoppe): Note that the triangle $$ABD$$ is isosceles, hence $$AD=BD$$.

Note that $$\sin 30^\circ=\frac{AC}{AB}=\frac12$$. From the Angle Bisector Theorem we find: $$\frac{AC}{AB}=\frac{CD}{BD} \Rightarrow CD=0.5BD \Rightarrow \frac{AD}{BC}=\frac{BD}{BD+0.5BD}=\frac23.$$

Method $$2$$ (Michael Rozenberg): Note that the triangle $$ABD$$ is isosceles, hence $$AD=BD$$.

Note that $$\sin 30^\circ=\frac{CD}{AD}=\frac12 \Rightarrow CD=0.5AD$$. Thus: $$\frac{AD}{BC}=\frac{AD}{CD+BD}=\frac{AD}{0.5AD+AD}=\frac23$$.

Method $$3$$: Refer to the figure:

$$\hspace{5cm}$$

Note that the triangle $$ABD$$ is isosceles, hence $$AD=BD$$.

Also, note that the triangles $$ACD, ADE, BDE$$ are congruent.

Let $$S$$ be the area of $$ABC$$. Then: $$\frac12\cdot AD\cdot AC\cdot \sin 30^\circ = \frac S3 \Rightarrow AD\cdot AC=\frac{4S}3;\\ \frac12\cdot BC\cdot \underbrace{AB}_{2AC}\cdot \sin 30^\circ = S \Rightarrow BC\cdot AC=2S;\\ \frac{AD}{BC}=\frac23.$$

Method $$4$$: Use the angle bisector formula $$l_a^2=\frac{bc}{(b+c)^2}\cdot [(b+c)^2-a^2]$$: \begin{align}AD^2&=\frac{AC\cdot AB}{(AC+AB)^2}\cdot [(AC+AB)^2-BC^2]=\\ &=\frac{AC\cdot 2AC}{(AC+2AC)^2}\cdot [(AC+2AC)^2-BC^2]=\\ &=\frac{2}{9}\cdot [(3\cdot \frac{BC}{\sqrt{3}})^2-BC^2]=\\ &=\frac{4}{9}BC^2 \Rightarrow \\ \frac{AD}{BC}&=\frac23.\end{align} Good luck on your studies!

Method $$5$$ (thinking out of box): (Michael Hoppe): Reflect the triangle on $$BC$$ as shown on the figure below.

$$\hspace{5cm}$$

Note that $$BC$$ and $$AD'$$ are both medians and at the intersection point $$D$$ they are divided in the ratio $$1:2$$, that is, $$CD:BD=1:2$$, which implies $$AD:BC=BD:BC=2:3$$.

Method $$6$$ (thinking out of box): Extend the side $$BC$$ such that $$CD=CD'$$ as shown on the figure below:

$$\hspace{5cm}$$

Note that the triangle $$ADD'$$ is equilateral triangle. Hence, $$AD=DD'$$, which are the radii of the circumscribed circle, implying $$DD'=BD$$. Hence, $$AD:BC=BD:BC=2:3$$.

• There is yet another solution, which requires no calculation at all; see my edit, please. Oct 28, 2018 at 18:29
• @MichaelHoppe, elegant solution of yours, indeed and challenge is accepted. Oct 29, 2018 at 4:03
• Method 6 is cute as well. Oct 29, 2018 at 10:50

$$\measuredangle C=180^{\circ}-30^{\circ}-60^{\circ}=90^{\circ}.$$ $$\measuredangle B=\measuredangle DAB=30^{\circ},$$ which gives $$BD=AD.$$

Also, since $$\measuredangle DAC=30^{\circ},$$ we obtain $$DC=\frac{1}{2}AD.$$

Id est, $$\frac{AD}{BC}=\frac{AD}{BD+DC}=\frac{AD}{AD+\frac{1}{2}AD}=\frac{2}{3}.$$