# Calculate the value of angle $ACB$

In an ABC triangle the angle $$BAC$$ is twice the angle $$ACB.$$ Consider a point $$D$$ in segment $$AC$$ so that the angle $$DBA$$ is twice the angle $$BAD.$$ Calculate the value of angle $$ACB,$$ knowing that the measurement of segment $$CD$$ is equal to the sum between twice the measurement of segment $$BD$$ and the length of segment $$AD.$$

Attemp:After using the law of sines on triangles ABD and BCD, I got the weird-looking equation attached. I think my approach most likely is not correct. $$4 \sin x \cos 2x= \sin(180 - 7x)$$

Denoting $$\angle ACB=\gamma$$ (the one we need to find), you have that $$CD=2BD+AD\Leftrightarrow \\ \frac{CD}{BD}=2+\frac{AD}{BD}\quad (1)$$ Apply sine law in $$\triangle ADB$$ and $$\triangle CDB$$: $$\frac{CD}{BD}=\frac{\sin7\gamma}{\sin\gamma}\\ \frac{AD}{BD}=\frac{\sin4\gamma}{\sin2\gamma}=2\cos2\gamma$$ so let $$\sin\gamma=x$$ and substitute in $$(1)$$. $$\cos2\gamma=1-2x^2$$ and $$\sin7\gamma=7x-56x^3+112x^5-64x^7$$ (see here). You get the equation $$64x^6-112x^4+52x^2-3=0\Leftrightarrow \\ (4x^2-3)(16x^4-16x^2+1)=0$$ which is solvable by letting $$t=x^2$$. And you have to take into account that $$7\gamma<180^{\circ}$$, so $$0. We get $$t=\frac{2-\sqrt{3}}{4}\Rightarrow x=\frac{\sqrt{2-\sqrt{3}}}{2} \Rightarrow \gamma=15^{\circ}$$ (see here for a table of trig. values).

• bjorn93. Great job. I found the same value by accurate drawing . Dec 20 '19 at 18:05

An euclidean-trigonometric cocktail

Take on $$DC$$ a point $$E$$ such that $$ED\cong BD$$. By our hypotheses (that can be rewritten as $$CD - BD \cong AD + BD$$) we have that $$CE\cong AE.$$ Produce $$AB$$ to $$F$$ so that $$\triangle AEF$$ is isososceles.

Call now $$\angle CAB = x$$, for simplicity. Then of course $$\angle ABD = 2x$$ and $$\angle ACB = \frac{x}2$$.

1. $$A$$, $$C$$, and $$F$$ lie on the circle centered in $$E$$ with radius $$\frac{\overline{AC}}2$$, therefore $$AF\perp CF$$.
2. Angle chasing yields $$\angle BEF \cong\angle EBF=\frac{\pi}2-\frac{x}2$$, therefore $$BF\cong \frac{AC}2$$.
3. External angle theorem yields $$\angle CBF = \frac{3x}2$$.

We must have $$\overline{AC} \sin x \cot \frac{3x}2 = \frac{\overline{AC}}2.$$ Trigonometric manipulation gives: $$\begin{eqnarray} \frac{\cos\frac{3x}2\sin x}{\sin\frac{3x}2} &=& \frac12\\ 2\frac{\cos \frac{x}2\cos x - \sin\frac{x}2\sin x}{\sin \frac{x}2\cos x + \cos\frac{x}2\sin x}\sin\frac{x}2\cos\frac{x}2 &=& \frac12\\ 2\frac{\cos\frac{x}2\left(2\cos^2\frac{x}2-1\right)-2\sin^2\frac{x}2\cos\frac{x}2}{\sin\frac{x}2\left(2\cos^2\frac{x}2-1\right)+2\cos^2\frac{x}2\sin\frac{x}2}\sin\frac{x}2\cos\frac{x}2 &=& \frac12\\ \frac{2\cos^2\frac{x}2\left(2\cos x-1\right)}{2\cos x +1}&=&\frac12\\ \frac{(\cos x+1)(2\cos x-1)}{2\cos x + 1}&=&\frac12, \end{eqnarray}$$ which in turns yields $$4\cos^2x - 3 = 0,$$ and thus $$x = \frac{\pi}6$$ as the only geometrically valid solution to the problem.

• $\angle EDB=3\alpha$ by external angle theorem? Dec 20 '19 at 18:08
• @bjorn93 yes, so $\angle EDB=\angle DAB+\angle DBA=3\alpha$ Dec 20 '19 at 18:19
• @bjorn93, thanks! I'll modify the path!
– dfnu
Dec 20 '19 at 18:19
• @bjorn93 thanks again for your observations! I revisited the solution with some trigonometric part, too.
– dfnu
Dec 20 '19 at 21:13

A purely euclidean path

Here is an approach based solely on congruences. It shares part of the path shown in my previous answer. I will repeat all the steps, anyway, to make this answer self-standing. Let $$\angle CAB = \alpha$$, so that $$\angle ABD = 2\alpha$$ and $$\angle ACB = \frac{\alpha}2$$.

1. Draw $$E$$ on $$CD$$ so that $$BD\cong DE$$; the hypothesis $$CD \cong 2BD + AD$$ implies $$AE\cong EC.$$
2. Produce $$AB$$ to $$F$$ so that $$AE\cong EF.$$Since $$A$$, $$C$$, and $$F$$ lie on the half-circle centered in $$E$$ and with radius $$\frac{\overline{AC}}2$$, we have $$CF\perp AF$$. Produce $$CF$$ to $$L$$, so that $$CF\cong FL$$.
3. Taking advantage of the fact that $$\triangle BDE$$ and $$\triangle ECF$$ are isosceles we obtain that $$\angle EBF \cong \angle BEF = \frac{\pi}2-\frac{\alpha}2$$. So $$BF \cong EF.$$
4. $$\angle BCF = \frac{\pi}2-\frac{3\alpha}2$$, and $$\angle BEC = \frac{\pi}2+\frac{3\alpha}2$$ by angle chasing.
5. $$\triangle BFC \cong \triangle BFL$$ by SAS criterion, implying in particular that $$\angle BLF \cong \angle BCF = \frac{\pi}2-\frac{3\alpha}2.$$
6. Points 4. and 5. imply that $$\square CEBL$$ is cyclic, because $$\angle BEC$$ and $$\angle BLC$$ are supplementary. Since $$AF \perp CL$$, and $$F$$ is the midpoint of $$CL$$, the center of its circumscribed circle must lie on $$AF$$. By 3., the center is $$F$$. So $$CF\cong \frac{AC}2$$ and the thesis, i.e. $$\boxed{\alpha = \frac{\pi}6},$$ follows immediately.