Let's break down and solve this geometry problem.

We are given:

• $AC$ and $CD$ are part of a straight line $ACD$.
• Triangle $ABC$ is isosceles with $AB = BC$.
• $\angle BCD = 138^\circ$, and $\angle ABC = 28^\circ$.

Step 1: Understanding the geometry

• Since triangle $ABC$ is isosceles, the two base angles $\angle BAC$ and $\angle BCA$ are equal.
• $\angle ABC = 28^\circ$ is given.
• We are tasked with finding the value of $x = \angle BAC$.

Step 2: Angle sum of a triangle

In triangle $ABC$, the sum of the interior angles is $180^\circ$. This gives us the following equation: $\angle ABC + \angle BAC + \angle BCA = 180^\circ$ Since $\angle BAC = \angle BCA$ (isosceles triangle), we can denote these angles as $x$. Thus, we have: $28^\circ + x + x = 180^\circ$ This simplifies to: $28^\circ + 2x = 180^\circ$ $2x = 180^\circ - 28^\circ = 152^\circ$ $x = \frac{152^\circ}{2} = 76^\circ$

Step 3: Using the straight line $ACD$

Since $ACD$ is a straight line, the angles $\angle BAC$ (or $x$) and $\angle BCD$ are supplementary: $x + 138^\circ = 180^\circ$ Substituting $x = 76^\circ$, we get: $76^\circ + 138^\circ = 214^\circ$ This is incorrect based on supplementary angles. Thus, checking the supplementary condition suggests an adjustment is needed. Based on the problem’s conditions, we cross-check for misinterpretations.