To solve the problem:

Given:

1. Triangles $\triangle ACD$ and $\triangle BCD$ are isosceles:
   • $AD \cong AC$
   • $BD \cong BC$
2. $\angle BAC = 18^\circ$
3. $\angle BDC = 48^\circ$
4. Find $\angle ABD$.

Solution:

Step 1: Consider $\triangle ACD$:

• Since $\triangle ACD$ is isosceles, $\angle CAD = \angle CDA$.
• The total angle sum of the triangle is $180^\circ$, so: $\angle ACD = 180^\circ - \angle BAC = 180^\circ - 18^\circ = 162^\circ.$ Therefore: $\angle CAD = \angle CDA = \frac{162^\circ}{2} = 81^\circ.$

Step 2: Consider $\triangle BCD$:

• Since $\triangle BCD$ is isosceles, $\angle CBD = \angle BCD$.
• The total angle sum of the triangle is $180^\circ$, so: $\angle BCD = 180^\circ - \angle BDC = 180^\circ - 48^\circ = 132^\circ.$ Therefore: $\angle CBD = \angle BCD = \frac{132^\circ}{2} = 66^\circ.$

Step 3: Calculate $\angle ABD$:

• The measure of $\angle ABD$ can be calculated as: $\angle ABD = \angle BAC + \angle CBD.$
• Substituting the given values: $\angle ABD = 18^\circ + 66^\circ = 84^\circ.$

Final Answer:

$\boxed{\angle ABD = 84^\circ.}$

Would you like further details or explanation?

Related Questions:

1. What is the measure of $\angle ACD$ in $\triangle ACD$?
2. Can you determine $\angle CAB$ if $\triangle BAC$ were equilateral?
3. How do you classify $\triangle ABD$ if all angles were known?
4. Why is $\triangle ACD$ isosceles important to the solution?
5. Can $\triangle BCD$ be solved using other properties, such as exterior angles?

Tip:

Always verify the sum of angles in a triangle equals $180^\circ$ to check your work!