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

Question

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?

Answer 1

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.

Answer 2

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

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

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

Answer 3

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:

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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:

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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!

Addendum.

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

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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:

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