Question: …ngle and an isosceles triangle share a common side. What is the measure of $\angle ABC$? $\angle ABC$ is $115^\circ$. Graph: top-right, a quadrilateral-like figure with vertices $A$ bottom-left, $B$ upper-left, $C$ upper-right, and $D$ lower-right; diagonal $BD$ is drawn. Tick marks indicate $AB = BD$, $BC = CD$, and a separate tick on the shared side/segment at the left. The angle at $A$ is marked $70^\circ$.

1. **Problem Statement:** We have a quadrilateral $ABCD$ with diagonal $BD$. It is given that $AB = BD$ and $BC = CD$, meaning triangles $ABD$ and $BCD$ are isosceles. The angle at $A$ is $70^\circ$. We need to find the measure of $\angle ABC$. 2. **Analyze the triangles:** - Since $AB = BD$, triangle $ABD$ is isosceles with $AB = BD$. - Since $BC = CD$, triangle $BCD$ is isosceles with $BC = CD$. 3. **Use the given angle:** $\angle BAD = 70^\circ$ (angle at $A$ in triangle $ABD$). 4. **Find angles in triangle $ABD$:** Since $AB = BD$, angles opposite these sides are equal: $$\angle ABD = \angle ADB = x$$ Sum of angles in triangle $ABD$: $$70^\circ + x + x = 180^\circ$$ $$2x = 110^\circ$$ $$x = 55^\circ$$ So, $$\angle ABD = 55^\circ$$ 5. **Find $\angle ABC$:** Note that $\angle ABC$ is the angle at $B$ in quadrilateral $ABCD$. It is composed of $\angle ABD$ (in triangle $ABD$) and $\angle DBC$ (in triangle $BCD$). 6. **Find $\angle DBC$ in triangle $BCD$:** Since $BC = CD$, triangle $BCD$ is isosceles with base $BD$. Let $\angle BCD = \angle CBD = y$. Sum of angles in triangle $BCD$: $$y + y + \angle BDC = 180^\circ$$ $$2y + \angle BDC = 180^\circ$$ 7. **Find $\angle BDC$:** $\angle BDC$ and $\angle ADB$ are adjacent angles on a straight line at point $D$. Since $\angle ADB = 55^\circ$ (from step 4), $$\angle BDC = 180^\circ - 55^\circ = 125^\circ$$ 8. **Calculate $y$:** $$2y + 125^\circ = 180^\circ$$ $$2y = 55^\circ$$ $$y = 27.5^\circ$$ 9. **Calculate $\angle ABC$:** $$\angle ABC = \angle ABD + \angle DBC = 55^\circ + 27.5^\circ = 82.5^\circ$$ **Final answer:** $$\boxed{82.5^\circ}$$