Coord-bashed this initially idk why ;-; . $\textbf{LEMMA: }$If $\sin(\theta_1) > \sin(\theta_2)$ and $0<\theta_1+\theta_2<180^{\circ},0<\theta_1<180^{\circ},0<\theta_2<180^{\circ}$, then $\theta_1>\theta_2$. $\textbf{PROOF: }$$\sin(\theta_1) > \sin(\theta_2)\implies\sin(\theta_1)-\sin(\theta_2)>0$ $\implies 2\cdot\cos\left(\dfrac{\theta_1+\theta_2}{2}\right)\cdot\sin\left(\dfrac{\theta_1-\theta_2}{2}\right)>0$. Now as $0^{\circ}<\dfrac{\theta_1+\theta_2}{2}<90^{\circ}$, we have $\cos\left(\dfrac{\theta_1+\theta_2}{2}\right)>0$. So from $2\cdot\cos\left(\dfrac{\theta_1+\theta_2}{2}\right)\cdot\sin\left(\dfrac{\theta_1-\theta_2}{2}\right)>0$, we get that $\sin\left(\dfrac{\theta_1-\theta_2}{2}\right)>0$. Now, $-90^{\circ}<\dfrac{-\theta_2}{2}<\dfrac{\theta_1-\theta_2}{2}<\dfrac{\theta_1}{2}<90^{\circ}\implies-90^{\circ}<\dfrac{\theta_1-\theta_2}{2}<90^{\circ}$. Also, as $\sin\left(\dfrac{\theta_1-\theta_2}{2}\right)>0$, we conclude $0^{\circ}<\dfrac{\theta_1-\theta_2}{2}<90^{\circ}\implies\dfrac{\theta_1-\theta_2}{2}>0^{\circ}\implies\theta_1>\theta_2$. Let $P=AB\cap CD$, $G=BD\cap AC$. Let $\angle DCG=\theta_1$ and $\angle GCB=\theta_2$ Now note that as $BC\parallel AD$ and $BC=\dfrac{AD}{2}$, we have that $BC$ is the $P-$midline of $\triangle PAD$. Thus $B$, $C$ become the midpoints of $PA$, $PD$ respectively. So, $G$ becomes the centroid of $\triangle PAD$. Now as centroid divides median in a $2:1$ ratio, we have by Ratio Lemma that $\dfrac{DG}{GB}=\dfrac{DC\cdot\sin(\theta_1)}{CB\cdot\sin(\theta_2)}\implies\dfrac{\sin(\theta_1)}{\sin(\theta_2)}=\dfrac{2}{1}\cdot\dfrac{CB}{DC}=2\cdot\dfrac{CB}{CP}=2\cdot\dfrac{\sin(\angle CPB)}{\sin(\angle CBP)}=2\cdot\dfrac{\sin\left(180^{\circ}-\angle PBC-\angle PCB\right)}{\sin(\angle PBC)}=2\cdot\dfrac{\sin\left(180^{\circ}-\angle A-\left(180^{\circ}-\dfrac{3\angle A}{2}\right)\right)}{\sin(\angle A)}=2\cdot\dfrac{\sin\left(\dfrac{\angle A}{2}\right)}{\sin(\angle A)}=2\cdot\dfrac{\sin\left(\dfrac{\angle A}{2}\right)}{2\cdot\sin\left(\dfrac{\angle A}{2}\right)\cdot\cos\left(\dfrac{\angle A}{2}\right)}=\dfrac{1}{\cos\left(\dfrac{\angle A}{2}\right)}$. Now as $0^{\circ}<\angle A<180^{\circ}$, we have $0<\cos\left(\dfrac{\angle A}{2}\right)<1\implies\dfrac{1}{\cos\left(\dfrac{\angle A}{2}\right)}>1$. So, $\dfrac{\sin(\theta_1)}{\sin(\theta_2)}=\dfrac{1}{\cos\left(\dfrac{\angle A}{2}\right)}>1\implies\sin(\theta_1)>\sin(\theta_2)$ and we are done using our Lemma as $0^{\circ}<\angle DCB=\theta_1+\theta_2<180^{\circ}$, $0^{\circ}<\theta_1<180^{\circ}$ and $0^{\circ}<\theta_2<180^{\circ}$. So, $\boxed{\angle ACD>\angle ACB}$. @3below just some nubmax things.....

It just suffices to compare $AD$ and $CD$.

This problem was such a nice trig bash, the first 3 lines of my solution were: me being me wrote: We first prove the following: Lemma: For any real $\theta \in (\pi/3]$, it holds that \[4\cos 2\theta + \frac{1}{\cos^2 \theta} - 1 < 4 \cos^2 \theta/2.\]

Why so complicated? Let $AB \cap CD=P$, just let $\angle BCA=\alpha$, now sine law in $\bigtriangleup PDA$ gives $\frac{AD}{CD}=\frac{2\sin \frac{\alpha}{2}}{\sin \alpha}$, so $CD=AD\cos \frac{\alpha}{2}$ so $CD<AD$, done. $\blacksquare$

nearly did the same thing but doesn't one and a half refers that the angles are in ratio 2:3..?