Answer:
The measure of angle ABC is;
[tex]90^{\circ}[/tex]
Explanation:
Given the figure in the attached image.
[tex]\begin{gathered} AB=CD \\ BC=AD \\ \text{ since} \\ AB=BC=15 \\ So; \\ AB=BC=CD=AD \end{gathered}[/tex]
Also;
Triangle BCD is an isosceles triangle;
[tex]\begin{gathered} 2\times\measuredangle\text{CBD}+90=180 \\ \measuredangle\text{CBD}=\frac{90}{2}=45^{\circ} \end{gathered}[/tex]
Then we have;
[tex]\begin{gathered} m\measuredangle ABC=m\measuredangle ABD+m\measuredangle CBD \\ m\measuredangle ABC=45^{\circ}+45^{\circ^{}} \\ m\measuredangle ABC=90^{\circ} \end{gathered}[/tex]
Therefore, the measure of angle ABC is;
[tex]90^{\circ}[/tex]
