Solution:

$\widehat{BAC}= 120^\circ$

Step by step solution:

Let $E$ be the symmetry point of $A$ through $D$

We get $ABEC$ is a parallelogram

Then: $\begin{cases}BE = AC = 4\ cm\\AE = 2AD = 2\sqrt3\ cm\end{cases}$

We have: $4^2 = 2^2 + \left(2\sqrt3\right)^2$

Or: $BE^2 = AB^2 + AD^2$

So that: $ABE$ is a right triangle (The converse of Pythagoras' Theorem)

Therefore: $\widehat{BAE}= \widehat{BAD}= 90^\circ$

Let $F$ be the midpoint of $AC$

and $G$ be the intersection point of $AD$ and $BF$

Then $G$ is centroid of triangle $ABC$

We get: $AG =\dfrac23AD = \dfrac{2}{\sqrt3}\ cm$

By applying $Pythagoras' \ Theorem$ in right triangle $ABG$, we get:

$\quad BG^2 = AB^2 + AG^2$

Or: $BG^2 = 4 + \dfrac43 =\dfrac{16}{3}$

So that: $BG = \dfrac{4}{\sqrt3}\ cm$

Recognize that $BG = 2AG$

Therefore $ABG$ is a half of equilateral triangle

It means: $\begin{cases}\widehat{AGB}= 60^\circ\\ \widehat{ABG}= 30^\circ\end{cases}$

Beside: $\triangle ABF$ is a isosceles triangle $(AB = AF = 2\ cm)$

We get: $\widehat{BAF}=180^\circ - 2\widehat{ABF}= 180^\circ - 2.30^\circ = 120^\circ$

Or: $\widehat{BAC}= 120^\circ$

Answer: $\widehat{BAC}= 120^\circ$