Finding an angle in a triangle, given the angle bisector and some conditions.
$ABC$ 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.
5 Answers
$\begingroup$A simple solution: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 $2\alpha=72^o$
$\endgroup$ 1 $\begingroup$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 $3x + 2\theta=180^o$
$\endgroup$ 1 $\begingroup$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?
$\endgroup$ $\begingroup$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 $\angle BAC$
$\endgroup$ $\begingroup$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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