See below in acute triangle $ABC$, $D$ is on $AC$ such that $AD=BC$. $CF$ is the angle bisector of $\angle ACB$. $DE \parallel CF$. $E$ is on $AB$. $AE = CD$.
Prove that $\angle ABC = 2 \angle BAC$. In addition, find the condition in which $\angle ADB = 3 \angle BAC$
I struggle to find ways to use two conditions $AD = BC$ and $AE = CD$
On
Here I prove that $\widehat{ABC} = 2\widehat{BAC}$ . Please note that the figure below is not precise.
From $B$ draw $BM$ perpendicular to $CF$ and extend it to meet $AC$ at point $N$ . Because $CM$ is the bisector of angle $\widehat C$ and perpendicular to $BN$ , triangle $CBN$ is isosceles and $BC = NC$ . The problems states that $BC = AD$ . Therefore $AD = NC$ , which in turn leads to $$AN = DC \qquad(1)$$
From $E$ draw a line parallel to $AC$ and let it meet $CF$ at point $O$. The quadrilateral $CDEO$ is a parallelogram because it has two pairs of parallel sides. We have $DC = EO$ . But the problem states that $DC = AE$ , therefore $$AE = EO = DC \qquad(2)$$ From (1) and (2) we have $AN = EO$ . The quadrilateral $AEON$ has a pair of parallel sides of equal length ($AN$ and $EO$) therefore it is a parallelogram. But from (2) we know that three sides of this parallelogram are equal, so it must be a diamond, and $$AE = EO = AN = NO \qquad(3)$$ Moreover, because $AEON$ is a diamond, the diagonal $AO$ bisects angle $\widehat A$. Now, $O$ is the meeting point of bisectors of $\widehat A$ and $\widehat C$ , therefore $BO$ is the bisector of $\widehat B$ .
Next, consider the triangle $OBN$ . Because $OM$ bisects $BN$ and is perpendicular to it, the triangle is isosceles and $$BO = NO \qquad(4)$$ From (3) and (4) we have $BO = EO$ , therefore the triangle $OBE$ is also isosceles and $\widehat{OBE} = \widehat{OEB}$ . But because $EO$ and $AC$ are parallel, $\widehat{OEB} = \widehat{BAC}$ . Therefore $$\widehat{OBE} = \frac12 \widehat{ABC} = \widehat{BAC}$$
Update: Here I prove that the statement $\widehat{ADB} = 3\widehat{BAC}$ which was initially written in the problem is not valid in general, and therefore cannot be proven. In fact, I demonstrate that the above statement is valid only for a particular class of similar triangles, and I specify the three angles of triangles of that class.
We previously saw that given the conditions of the problem, we necessarily have $\widehat{ABC} = 2 \widehat{BAC}$ . Therefore in triangle $ABC$ : $$\widehat{ACB} = 180^o - 3 \widehat{BAC} $$ Now let us assume that the construction is such that $\widehat{ADB} = 3\widehat{BAC}$ . This means that $$\widehat{BDC} = 180^o - \widehat{ADB} = 180^o - 3 \widehat{BAC}$$ which, in turn means that the triangle $BDC$ must be isosceles for this condition to hold. So $$BC = BD$$ Recall that the problem states that $$BC = AD$$ Therefore: $$BD = AD$$ which means the triangle $ADB$ must also be isosceles. Therefore $$\widehat{ABD} = \widehat{BAD} $$ By writing the sum of angles in triangle $ABD$ we are able to find the angle $\widehat{BAC}$ : $$\widehat{ABD} + \widehat{BAD} + \widehat{ADB} = 5\widehat{BAD} = 180^o$$ $$\widehat{BAD} = \widehat{BAC} = \frac{180^o}{5} = 36^o $$ So, given the conditions of the problem, the statement $\widehat{ADB} = 3 \widehat{BAC}$ is valid only if angle $\widehat A$ in triangle $ABC$ be $36^o$ . Hence that statement is not valid for other triangles. It follows that in the same triangle $ABC$, angle $\widehat B$ will be $2 \times 36^o = 72^o$ and $\widehat C$ will be $180^o - (36^o + 72^o) = 72^o$. So the triangle $ABC$ will be isosceles. Finally, note also that in this case $\widehat{ABD} = \widehat{DBC} = 36^o$ , which means $BD$ bisects angle $\widehat B$ in triangle $ABC$ , so $BD$ passes through point $O$ and in the above figure, $D$ falls on $G$ .
Not an elegant solution, because I'm using trigonometry, but just to show that the correct implication is $∠=2∠$.
Set $a=AE$, $b=AD$ and $x=a/b$. Note first of all that from similarity of triangles $AED$, $AFC$ one gets: $$ EF={a^2\over b}. $$ Moreover, from the angle bisector theorem in triangle $ABC$ we also get: $$ BF=a. $$ If $\alpha=∠$ from the cosine rule we obtain then: $$ \cos\alpha={2+5x+4x^2+x^3\over4+6x+2x^2}={x+1\over2}. $$ And if $\beta=∠$ from the same rule we get: $$ \cos\beta={-2+3x+4x^2+x^3\over4+2x}={x^2+2x-1\over2} =2\cos^2\alpha-1=\cos2\alpha. $$
EDIT.
See diagram below to get convinced that $\angle ADB$ is NOT in general equal to $3\alpha$.