Prove that AC = BD and AD = BC from the given figure.

From the figure, we can observe:

• $AO = BO$ (indicated by the single tick marks on segments $AO$ and $BO$).
• $CO = DO$ (indicated by the double tick marks on segments $CO$ and $DO$).
• $\angle AOC$ and $\angle BOD$ are vertically opposite angles.

Consider $\triangle AOC$ and $\triangle BOD$:

1. $AO = BO$ (Given from the figure)
2. $\angle AOC = \angle BOD$ (Vertically opposite angles)
3. $CO = DO$ (Given from the figure)

By the Side-Angle-Side (SAS) congruence criterion, $\triangle AOC \cong \triangle BOD$.

Since $\triangle AOC \cong \triangle BOD$, their corresponding parts are equal:

• $AC = BD$ (Corresponding sides)
• $AD = AO + OD$ and $BC = BO + OC$. Since $AO = BO$ and $OD = OC$, it follows that $AD = BC$.

Consider $\triangle ACD$ and $\triangle BDC$:

1. $AC = BD$ (Proven in Step 3)
2. $CD = DC$ (Common side)
3. $AD = BC$ (Proven in Step 3)

By the Side-Side-Side (SSS) congruence criterion, $\triangle ACD \cong \triangle BDC$.

Since $\triangle ACD \cong \triangle BDC$, their corresponding angles are equal.
The angle $\angle ACD$ in $\triangle ACD$ corresponds to the angle $\angle BDC$ in $\triangle BDC$.
Therefore, $\angle ACD = \angle BDC$.

This completes the proof.

$\boxed{\text{Proven that } \angle ACD = \angle BDC}$