Deltoid Geometry: Calculate the 1:3 Point Division and Triangle Area Ratio

To find the ratio of areas between $\triangle ACD$ and $\triangle BAD$, we start by examining the information given. The ratio $CK : AC = 1:3$ implies that $CK$ is one-third of $AC$.

The line segment $AC$ is divided into $CK$ and $AK$, with $AK = 2 \times CK$ due to $\frac{CK}{AC} = \frac{1}{3}$ and $\frac{AK}{AC} = \frac{2}{3}$. The triangles $\triangle ACK$ and $\triangle ACD$ share the same height from vertex $A$ to line $CD$.

Because the triangles share this common height, their areas are proportional to their respective base segments $CK$ and $AC$.

Thus, $\frac{\text{Area of } \triangle ACK}{\text{Area of } \triangle ACD} = \frac{CK}{AC} = \frac{1}{3}$.

Since $\triangle ACD = \triangle ACK + \triangle CKD$ and $\triangle BAD = \triangle ACK + \triangle CKD$, we look at the sums such that:

• The area of $\triangle ACD$ consists of the full base $AC$ and its corresponding height.
• The area of $\triangle ACK$ has a base of $CK$ and the same height.
• Given $\triangle ACK$ has 1 unit area for every 3 units of $\triangle ACD$, $\triangle ACD$'s area is 4 times that of $\triangle ACK$.

The ratio of the area of $\triangle ACD$ to $\triangle BAD$ becomes $\frac{\text{Area of } \triangle ACD}{\text{Area of } \triangle BAD} = \frac{1}{4}$. Thus, $\triangle ACD$ is 4 times smaller than $\triangle BAD$.

Therefore, the ratio of areas between $\triangle ACD$ and $\triangle BAD$ is $\boxed{1:4}$.