Let the foot of the altitudes from $D$ to $BC$ be $Q$, and let the midpoint of $AB$ be $M$. We have $BM=\frac{1}{2}AB=\frac{1}{2}CD=PD$. Since $BM||PD$, this implies that $BMPD$ is a rectangle, and $MD\perp BM$. We now see that $MD$ is the perpendicular bisector $AB$, so $AD=BD$. Finally, we have $BC=2AD=2BD$, and by Sine Law, we can get that $\angle BDC=90^\circ$, and therefore, $\angle DBC=60^\circ$. Hence, $\angle BAD=\angle ABD=90^\circ-\angle DBC=30^\circ$.

Construct the rectangle $ABCE$, see that $\triangle CDE$ is equilateral, while in $\triangle DAE, AE=2AD$ and $m(\widehat{AED})=30^\circ$, that is, $\widehat{DAE}=60^\circ$, or $\widehat{BAD}=30^\circ$. Best regards, sunken rock

This is the first time I was and probably last time I will be, able to solve a Geo doing a construction. Construct a rectangle $ABCD'$ such that $AB = CD'$ and $BC = AD'$ then by some simple angle chase we get $\angle DD'A = 30^o$. Let $\angle BAD = x \implies \angle DAD' = 90 - x$ and $\angle ADD' = 60 + x$. Since $BC = AD' = 2AD$ , Using sine law we get $$\frac{\text{sin}(60 + x)}{2AD} = \frac{\text{sin}(30^o)}{AD} \implies x = 30^o$$ since clearly $\angle BAD$ is acute.

Extend CD to meet AB at P So, angle CPB=60° Let AD=x, so, BC=2x In ∆ BCP, CP =2BP PD+CD=2BP PD+AB=2BP( CD=AB) PD+AP+PB=2BP AP+PD=BP In∆BCP, tan 60°=2x/BP BP/x=2/√3 Let angle BAD=y angle PDA=60°-y So, by laws of sine, PD+AP/ AD={sin y+ sin(60°-y)}/ sin 60° BP/x={sin y+ sin(60°-y)}/ sin 60° {sin y+ sin(60°-y)}/ sin 60°=2/√3 sin y +√3 cos y=2 4cos²y-4√3cos y+3=0 (2cos y-√3)²=0 cos y=√3/2 y=30° angle BAD=30° @Krishijivi