Let's break down the given information to find the angle $\angle CAB$ in triangle $ABC$:

Step-by-Step Analysis:

1. Triangle Setup:

   • Let $\triangle ABC$ be the triangle, with $D$ being the midpoint of side $AB$.
   • $\angle ADC = 45^\circ$.
   • $\angle DBC = 30^\circ$.

2. Important Relationships:

   • Since $D$ is the midpoint of $AB$, $AD = DB$.
   • $\angle DBC = 30^\circ$ is given, so $\angle DBA = 30^\circ$ because $D$ is the midpoint and forms two equal angles.

3. Angles in $\triangle BCD$:

   • $\angle DBC = 30^\circ$ and $\angle DCB = 45^\circ$ (because $\angle ADC = 45^\circ$ and $\angle DCB$ is supplementary to it in the line of triangle $ABC$).
   • Now, use the angle sum of a triangle in $\triangle BCD$ to find $\angle BCD$.

4. Using the Angle Sum of Triangle $ABC$:

   • With the above relationships, the angle $\angle CAB$ will be calculated using the sum of angles in triangle $ABC$ and additional geometric properties.

Let me calculate the precise result for $\angle CAB$ and finish the steps for you.To solve for $\angle CAB$, let’s proceed step by step:

Step 1: Labeling the Angles

• Given that $D$ is the midpoint of $AB$, we know that $AD = DB$.
• Therefore, triangle $ABD$ is an isosceles triangle, and the angles at $A$ and $B$ must be equal.
• Let $\angle DBC = 30^\circ$, and we also know that $\angle ADC = 45^\circ$.

Step 2: Angles in Triangle $DBC$

• $\angle DBC = 30^\circ$ and $\angle DCB = 45^\circ$.
• The sum of the angles in $\triangle DBC$ must be $180^\circ$.
• So, we can find $\angle BDC$:

$\angle BDC = 180^\circ - \angle DBC - \angle DCB = 180^\circ - 30^\circ - 45^\circ = 105^\circ.$

Step 3: Finding $\angle ABD$

• Now, notice that $\angle ABD$ is part of the triangle $ABD$. Since $AD = DB$ (midpoint), triangle $ABD$ is isosceles.
• Therefore, $\angle ABD = \angle ADB$.
• The sum of angles in triangle $ABD$ gives:

$\angle ABD + \angle ADB + \angle BAD = 180^\circ.$

Since $\angle ADB = 45^\circ$ (angle at $D$ is part of the configuration we were given), we can write:

$2 \times \angle ABD + 45^\circ = 180^\circ.$

Solving for $\angle ABD$:

$2 \times \angle ABD = 180^\circ - 45^\circ = 135^\circ,$

$\angle ABD = 67.5^\circ.$

Step 4: Finding $\angle CAB$

• $\angle CAB$ is the same as $\angle ABD$, because $\angle ABD$ and $\angle CAB$ are angles that share the same vertex $A$ in $\triangle ABC$.

Thus, $\angle CAB = 67.5^\circ$.

Final Answer:

$\boxed{67.5^\circ}$

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Would you like more details on any specific step, or have further questions? Here are some related questions you might find useful:

1. How can the concept of midpoints help in solving other types of geometric problems?
2. What is the relationship between angles in isosceles triangles?
3. How do you use the angle sum property in non-right-angled triangles?
4. What happens if the midpoint divides the triangle into right-angled triangles?
5. How can geometric constructions help verify angle measurements?

Tip: Always remember to check if a triangle has properties like isosceles or symmetric, as this can simplify your calculations!