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$ABC$ is a triangle in which $\angle B= 2\angle C$. $D$ is a point on side $BC$ such that $AD$ bisects $\angle BAC$ and $AB = CD$.

What is the measure of $\angle BAC? $

I tried using angle bisector theorem, similarity, but the sides, don't include the angles. Also I got a relation, $3x+2\theta=180^{\circ}$, where $\angle ACB=x;\angle BAD=\theta$.

But I cannot derive one more relation, please help.
Please don't use trigonometry, nor any constructions. We have to do without them, simply.

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  • $\begingroup$ Please I want the answer urgently. $\endgroup$ – Aditya Agarwal Sep 15 '15 at 14:14
  • $\begingroup$ Please can someone, just point me in the right direction? $\endgroup$ – Aditya Agarwal Sep 15 '15 at 14:14
  • $\begingroup$ It is not a good attitude to put us in a hurry. We give help because we like it, but you should avoid abusing it. $\endgroup$ – Jack D'Aurizio Sep 15 '15 at 14:35
  • $\begingroup$ Sorry, I am desperate right now. Thanks for the answer btw, but please see the edit. $\endgroup$ – Aditya Agarwal Sep 15 '15 at 14:35
  • $\begingroup$ I have already answered to that, too. You may check that if $D,E,F,G,H$ are the vertices of a regular pentagon, $DEG=ABC$ works just fine. $\endgroup$ – Jack D'Aurizio Sep 15 '15 at 14:38
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A simple solution: enter image description here Let $\angle ACB=\gamma$ and $\angle BAC=2\alpha$. By assumption $\angle ABC=2\gamma$

Let $BP$ is the bisector of $\angle ABC$ where $P$ is on $AC$. Then $\angle PCB=\angle PBC=\gamma$. This means $PC=PB\quad (1)$. But it is also given $CD=AB\quad (2)$.

From $(1),(2)\Rightarrow \triangle ABP \simeq \triangle DCP\,\,$ (they have a common angle) $\,\,\Rightarrow AP=DP$ $\Rightarrow \angle PAD=\angle PDA=\alpha=\angle BAD$ (because $AD$ is a bisector). Therefore $PD ||AB\Rightarrow \angle PDC=\angle ABC=2\gamma\quad (3)$. But from $\triangle ABP \simeq \triangle DCP$ we also have $2\alpha=\angle BAP=\angle CDP\quad (4)$.

Finally, $(3)$ and $(4)$ give $2\alpha=2\gamma$ and from $2\alpha+3\gamma=180^o$ we find $2\alpha=72^o$

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  • $\begingroup$ Glad that it helped you. $\endgroup$ – Svetoslav Sep 16 '15 at 12:46
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Hint:

Use $2$ times the $sin$-theorem respectively for triangle $ADC$ and $ABD$ and the bisector theorem : $AB:AC=BD:DC$ together with the equality $3x + 2\theta=180^o$

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  • $\begingroup$ Please can you restrict the use of trigonometry? $\endgroup$ – Aditya Agarwal Sep 15 '15 at 14:17
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Assume $R=1$. Then $AB=2\sin(\theta),AC=2\sin(2\theta)$ and $BC=2\sin(3\theta)$ by the sine theorem.

The bisector theorem gives: $$CD = 2\sin(3\theta)\frac{\sin(2\theta)}{\sin(2\theta)+\sin(\theta)} = 2\sin(\theta)=AB$$ hence we have: $$ \sin(3\theta)\sin(2\theta)=\sin(\theta)\sin(2\theta)+\sin^2(\theta) $$ or, by dividing both terms by $\sin^2(\theta)$, $$ (4\cos^2(\theta)-1)2\cos(\theta) = 2\cos(\theta)+1 $$ so: $$ 8\cos^3(\theta)-4\cos(\theta)-1=0$$ then $\cos(\theta)\in\left\{-\frac{1}{2},\frac{1-\sqrt{5}}{4},\frac{1+\sqrt{5}}{4}\right\}$ or $\theta\in\left\{\frac{2\pi}{3},\frac{\pi}{5},\frac{3\pi}{5}\right\}$, hence $\widehat{ACB}=\theta=\frac{\pi}{5}$ and $$\widehat{BAC}=\pi-3\theta=\color{red}{\frac{2\pi}{5}}=72^\circ=\widehat{ABC}.$$ Do you recognize a portion of a regular pentagon?

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Notice, $$\angle ADB=180^\circ -\left(\angle B+\frac{\angle BAC}{2}\right)$$ $$\angle B+\angle C+\angle BAC=180^\circ\iff 2\angle C+\angle C+\angle BAC=180^\circ$$ $$\iff \angle C=\frac{180^\circ-\angle BAC}{3}\tag 1$$$$\iff \angle B=\frac{2(180^\circ-\angle BAC)}{3}\tag 2$$

apply sine rule in $\triangle ABD$ $$\frac{\sin\angle B}{AD}=\frac{\sin \angle ADB}{AB}$$ $$\frac{\sin\angle B}{AD}=\frac{\sin \left(\angle B+\frac{\angle BAC}{2}\right) }{AB}\tag 3$$ Similarly, apply sine rule in $\triangle ADC$

$$\frac{\sin\angle C}{AD}=\frac{\sin \left(\frac{\angle BAC}{2}\right)}{CD}$$ Setting $CD=AB$, we get $$\frac{\sin\angle C}{AD}=\frac{\sin \left(\frac{\angle BAC}{2}\right)}{AB}\tag 4$$

Dividing (3) by (4), we get

$$\frac{\sin\angle B}{\sin \angle C}=\frac{\sin \left(\angle B+\frac{\angle BAC}{2}\right) }{\sin \left(\frac{\angle BAC}{2}\right) }$$ Setting the values of $\angle B$ & $\angle C$, we get $$\frac{\sin \left(\frac{2(180^\circ-\angle BAC)}{3}\right)}{\sin \left(\frac{180^\circ-\angle BAC}{3}\right)}=\frac{\sin \left(\frac{2(180^\circ-\angle BAC)}{3}+\frac{\angle BAC}{2}\right) }{\sin \left(\frac{\angle BAC}{2}\right) }$$

$$2\cos \left(60^\circ-\frac{\angle BAC}{3}\right)=\frac{\sin \left(120^\circ-\frac{\angle BAC}{6}\right) }{\sin \left(\frac{\angle BAC}{2}\right) }$$

I hope you can solve for $\angle BAC$

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The angle bisector $AD$ has length as a harmonic mean $ 2 b c/( b+c) $

$$ 2 \theta + 3 x = \pi $$

$$ \frac{\sin \theta }{\sin x}=\frac{ b+c }{2 b } $$

Please take it further to find two unknowns from two equations above.

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