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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. $$4 \sin x \cos 2x= \sin(180 - 7x)$$

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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 $0<x<\sin\frac{180^{\circ}}{7}<\sin\frac{180^{\circ}}{6}=\frac 12\Rightarrow 0<t<\frac 14$. 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).

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  • $\begingroup$ bjorn93. Great job. I found the same value by accurate drawing . $\endgroup$
    – sirous
    Dec 20 '19 at 18:05
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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 $\angle CBF = \frac{3x}2$.

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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.

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  • $\begingroup$ $\angle EDB=3\alpha$ by external angle theorem? $\endgroup$
    – bjorn93
    Dec 20 '19 at 18:08
  • $\begingroup$ @bjorn93 yes, so $\angle EDB=\angle DAB+\angle DBA=3\alpha$ $\endgroup$
    – bjorn93
    Dec 20 '19 at 18:19
  • $\begingroup$ @bjorn93, thanks! I'll modify the path! $\endgroup$
    – dfnu
    Dec 20 '19 at 18:19
  • $\begingroup$ @bjorn93 thanks again for your observations! I revisited the solution with some trigonometric part, too. $\endgroup$
    – dfnu
    Dec 20 '19 at 21:13
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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 $\angle ACB = \frac{\alpha}2$.

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  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.
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