Let $BE$ and $CF$ be the altitudes from $B$ and $C$, respectively, of the triangle $ABC$, and denote by $H$ their meeting point. The three perpendiculars theorem implies that $DE \perp AC$ and $DF \perp AB$.
Unfold the tetrahedron on the plane $(ABC)$ and denote by $D_1$ the image of the vertex $D$ of the triangle $DAB$ and by $D_2$ the image of the vertex $D$ of the triangle $DAC$, after the unfolding.

The perpendicularities $DE \perp AC$ and $DF \perp AB$ are maintained on the unfolding. Therefore, we deduce that $D_1, F$ and $C$ are collinear. Similarly, we find that $D_2, E$ and $B$ are collinear.

As $\widehat{BAC} + \widehat{CAD} + \widehat{DAB} = 180^\circ$, the points $D_1, A$ and $D_2$ are collinear; obviously, $A$ is the midpoint of the segment $D_1D_2$.

$\widehat{ABC} \equiv \widehat{DAB}$, hence the lines $D_1D_2$ are $BC$ parallel, so $D_1D_2CB$ is a trapezium, whose diagonals meet in $H$. The midpoint of the larger base lies on the line $AH$, therefore $AH$ also contains the midpoint of the smaller base $BC$. Because $AH$ is simultaneously an altitude and a median of the triangle $ABC$, it follows that the triangle $ABC$ is isosceles, with $AB = AC$.

The line $AH$ is the common perpendicular bisector of the bases of the trapezium $D_1D_2CB$, therefore $D_1D_2CB$ is an isosceles trapezium, hence $D_1B = D_2C$ and thus $DB = DC$.

Alternative solution:

From $\widehat{BAC} + \widehat{CAD} + \widehat{DAB} = 180^\circ$, $\widehat{ABC} \equiv \widehat{DAB}$ and $\widehat{BAC} + \widehat{ABC} + \widehat{ACB} = 180^\circ$, we deduce that $\widehat{CAD} \equiv \widehat{ACB}$.

Let $H$ be the orthocenter of the triangle $ABC$ and $BE, CF, AG$ its altitudes. From the three perpendiculars theorem, we deduce $DE \perp AC, DF \perp AB$ and $DG \perp BC$.

Triangles $ABG$ and $DAF$ are similar, therefore we have $AG \cdot AD = DF \cdot AB$. Moreover, from the similarity of the triangles $ACG$ and $DAE$ we deduce $AG \cdot AD = DE \cdot AC$. We obtain $DF \cdot AB = DE \cdot AC$, and thus $A_{ABD} = A_{ACD}$.

From $AG \perp BC$ and $DG \perp BC$ we deduce $BC \perp (ADG)$, whence $BC \perp AD$. If $BI \perp AD, I \in AD$, then $AD \perp (BIC)$.
We obtain $CI \perp AD$, therefore $A_{ABD} = A_{ACD}$ iff $BI = CI$, thus the triangles $AIB$ and $AIC$ are congruent, whence we have $AB = AC$. Similarly, we deduce that the triangles $DIB$ and $DIC$ are congruent, therefore $DB = DC$.