lasha wrote: In triangle $ ABC$ we have $ \angle{ACB} = 2\angle{ABC}$ and there exists the point $ D$ inside the triangle such that $ AD = AC$ and $ DB = DC$. Prove that $ \angle{BAC} = 3\angle{BAD}$. Let's $ \angle ABC=\beta$, $ \angle CBD = \theta$ So we have: $ \angle ABD = \beta - \theta$ $ \angle ACD = 2\beta - \theta$ $ \frac{BC}{CD}=2\cos \theta$ $ \frac{CD}{AC}=2\cos (2\beta - \theta)$ therefore $ \frac{\sin 3\beta}{\sin \beta}=\frac{BC}{AC}=4\cos \theta \cos (2\beta - \theta)=2cos(2\beta-2\theta)+2cos2 \beta$ $ \leftrightarrow$ $ \sin \beta \cos 2\beta + \sin 2\beta cos \beta=2sin \beta cos (2\beta-2 \theta) + 2sin\beta cos 2\beta$ $ \leftrightarrow$ $ \sin \beta =2 sin \beta \cos(2\beta -2\theta)$ $ \leftrightarrow$ $ cos(2\beta -2\theta)=\frac{1}{2}$ $ \leftrightarrow$ $ \angle ABD=\beta-\theta=30^o$ We also have that: $ \angle BAD=\beta - 2\theta$ $ \angle CAD=180^o+2\theta -4\beta$ So it's left to prove that $ 2\angle BAD=2\beta - 4\theta=180^o+2\theta -4\beta=\angle CAD$ or $ \beta -\theta = 30^o$ which we already had proved.

Can someone prove this without trigonometry?

Let L be the reflection of D wrt AB then $ \angle ADL = 2\angle ADI$ So we need to prove that $ \angle DAL = \angle DAC or DL = DC = DB$ But $ DB = LB$ therefore $ DLB$ is equilateral triangle. The mean idea is to show that $ \angle DBA = 30^o$. Construct C-bisector of triangle ABC, it intersects AB at I. Since $ \angle ACB = 2\angle ABC$ then $ IC = IB$ which follows that $ ID$ is perpendicular bisector of segment $ BC$. We have $ AI.AB = AC^2 = AD^2$ then $ \angle ADI = \angle DBI$ Denote $ M$ lie on $ (A, AC)$ such that $ AM//BC. AM\cap AB = \{I'\}$ Applying Thales's theorem we get $ \frac {AM}{BC} = \frac {AI'}{I'B} = \frac {AC}{BC}$ so $ I'\equiv I$. It's easy to see $ IA = IM$ and $ ID\perp AM$ thus $ AD = DM = AM$ Then $ ADM$ is equilateral triangle. We obtain $ \angle ADI = 30^o = \angle DBA$ Our proof is completed.

Attachments:
picture22.pdf (9kb)

It becomes so easy after somebody else made it! Take D’ symmetrical of D about AB, see that $ \triangle BDD'$ is equilateral, hence DD’ = DC and triangles ADD’ and ADC are congruent. Best regards, sunken rock

Draw the $\perp $ bisector of $BC $ through $D $ to meet $AB $ at $M $ and $BC$ at $N $. Observe that $\angle ABC = \angle MCB = \angle ACM = x $ (say). So, $AD^2 = AC^2 = AM\cdot AB $. This gives $\frac {DA}{DB} = \frac {MA}{MD}$, or, $\frac {AC}{DC} = \frac {MA}{MD}$, or, $\frac {AC}{AM} = \frac {CD}{DM}$, or, $\frac {BC}{BM} = \frac {CD}{DM} $, or, $\frac {\sin 2\angle x}{\sin \angle x} = \frac {\cos \angle x}{\sin \angle ABD} $. Thus, $\sin \angle ABD = \frac {1}{2} $. So, $\angle ABD$ = $30$°. Now, let $P $ be the foot of the $\perp $ from $A $ to $CD $. Observe that $DP = \frac{DB}{2} $. Applying sine law in triangles $ADP $ and $ADB $, we get $\angle BAD = \angle PAD $. The result now follows immediately.

Dear MLs, see attachment. Mid-perpendicular of BC is the symmetry axis. AM // BC , AMD is equilateral Now angle equalities are easily seen, Required angle relation is clear. M.T.

Attachments:
ML262463.doc (29kb)

Firstly note that $D$ is a well defined point inside the triangle given by the intersection of the perpendicular bisector of $BC$ with the circle with center $A$ and radius $AC$. Thus, our configuration is unique. Now, assume that $\angle ABC= \beta$ and $\angle DBC= x$. Then simple angle chasing gives: $$\angle DAB=\beta -2x, \angle DAC= \pi-4\beta+2x, \angle ABD=\beta-x, \angle DBC=\angle DCB=x, \angle DCA= 2\beta-x $$ Now, the cevians $AD,BD$ and $CD$ are concurrent. Hence, by the trigonometric form of Ceva's theorem, $$\frac{\sin(\beta-2x)}{\sin(\pi-4\beta+2x)}.\frac{\sin(x)}{\sin(\beta-x)}.\frac{\sin(2\beta-x)}{\sin(x)}=1$$$$\Rightarrow \sin(\beta-2x)=2\sin(\beta-x)\cos(2\beta-x)$$where we used the facts that $\sin(\pi-x)=\sin(x)$ and $\sin (2\alpha)=2\sin \alpha \cos \alpha$. Now, it is easy to see that $x=\beta-\frac{\pi}{6}$ is 'a' solution of this equation. But since our configuration is unique, hence $x=\beta-\frac{\pi}{6}$ is 'the' value of $x$. Thus $$\angle BAD=\beta-2x=\frac{\pi}{3}-\beta=\frac{1}{3}(\pi-3\beta)=\frac{1}{3}\angle BAC$$.