In triangle $ABC$ find angle $\angle BAC$ given that...

Question

In triangle $ABC$, $D$ is a point on $BC$ such that $CD=2BD$.Also we know that $\angle B=45$ and $\angle ADC=60$.Find angle $\angle BAC$.

I tried to find two similar triangles but failed! I heared the aimed solution method uses similar triangles...

Answer 1

Let $E$ be the orthogonal projection of vertex $C$ onto $AD$ so $CE \perp AD$. Then $CDE$ right-angled with $CDE =\angle \, 60^{\circ}$. Therefore, $$ED = \frac{1}{2} \, CD = BD$$ and $$\angle \, BCE = \angle \, DCE = 30^{\circ}$$ Consequently, triangle $BDE$ is isosceles so $\angle \, DBE = \angle \, DEB = 30^{\circ}$.

Therefore $\angle \,CBD = \angle \, DBE = 30^{\circ} = \angle \, BCE$ which means that triangle $BCE$ is also isosceles. Hence $$BE = CE$$ Now, after calculating that $$\angle \, ABE = ABC - \angle \, CBE = 45^{\circ} - 30^{\circ} = 15^{\circ} = \angle \, BAE$$ triangle $ABE$ is also isosceles, yielding $$AE = BE$$ Therefore $$AE = CE$$ which means that triangle $ACE$ is isosceles with $\angle \, AEC = 90^{\circ}$. Therefore $\angle \, EAC = 45^{\circ}$. Thus $$\angle \, BAC = \angle \, BAE + \angle \, EAC = 15^{\circ} + 45^{\circ} = 60^{\circ}$$

Answer 2

From triangle $ABD$, $\frac{1}{\sin 15^\circ} = \frac{AD}{\sin 45^\circ}$ and from triangle $ADC$, we have $\frac{AD}{\sin (120^\circ-\theta)} = \frac{2}{\sin\theta}$ where $\theta = \angle DAC$. Substituting for $AD$ in the second from the first, we get $$2\sin 15^\circ \sin(120^\circ-\theta) = \sin 45^\circ \sin \theta$$ This gives $\theta = 45^\circ$. Hence $\angle BAC = 60^\circ$.

Answer 3

The solution I am going to give is not shorter than yours but is useful (I think so) in that it presents a general methodology for solving this type of problem, i.e., systematically using $\tan$ function, and basic algebra).

Let $H$ be the foot of the altitude from $A$ on side $BC$.

As the problem is up to a scaling, we can assume that $AH=1.$

Let $CH=x$, $HD=y$ and $DB=z$.

We can build a system of 3 equations with these 3 unknowns:

$y+x=2z, \ \ \dfrac{1}{x}=\sqrt{3} \ \ and \ \ \ \dfrac{1}{x+z}=1 \ $

(the last ones coming from the computation of certain tangents.)

which gives easily $x,y,z$ (up to you...), from which $\tan(\widehat{ABH})=\dfrac{1}{y}$ giving $\widehat{ABH}$.

Then $\widehat{CAB}=180°-(\widehat{ABH}+\widehat{CBA})=60°$.

Answer 4

Let $E$ be the projection of $A$ on $BC$ and let us assume $AE=\sqrt{3}$.
We have $ED=1$ and $EB=\sqrt{3}$, from which $DB=\sqrt{3}-1$ and $CE=2\sqrt{3}-3$.
By the Pythagorean theorem, $AB=\sqrt{6}$ and $AC=2\sqrt{6-3\sqrt{3}}=\sqrt{6}(\sqrt{3}-1)$.
By the cosine theorem:

$$\cos\widehat{BAC}=\frac{AB^2+AC^2-BC^2}{2\,AB\cdot AC} = \frac{6+12(2-\sqrt{3})-18(2-\sqrt{3})}{12(\sqrt{3}-1)}=\frac{1}{2}$$ hence $\widehat{BAC}=\color{red}{60^\circ}$ as wanted.