Isosceles Triangle Angle 4C1272

1. **Problem Statement:** We have a quadrilateral with points $A$, $B$, $C$, and $D$. The segments $AB$, $BD$, $BC$, and $CD$ are marked equal, and $\angle A = 70^\circ$. We want to find the measure of $\angle ABC$. 2. **Given Information:** - $AB = BD = BC = CD$ - $\angle A = 70^\circ$ - $\triangle ABD$ is isosceles with $AB = BD$. - $\triangle BCD$ is equilateral with $BC = CD = BD$. 3. **Step 1: Analyze $\triangle BCD$ (Equilateral Triangle)** Since $BC = CD = BD$, $\triangle BCD$ is equilateral, so all its angles are $60^\circ$. Therefore, $\angle BCD = 60^\circ$ and $\angle CBD = 60^\circ$. 4. **Step 2: Analyze $\triangle ABD$ (Isosceles Triangle)** Since $AB = BD$, $\triangle ABD$ is isosceles with base $AD$. We know $\angle A = 70^\circ$. 5. **Step 3: Find $\angle ABD$ in $\triangle ABD$** Sum of angles in $\triangle ABD$ is $180^\circ$: $$\angle A + \angle ABD + \angle BDA = 180^\circ$$ Since $AB = BD$, $\angle ABD = \angle BDA$. Let this common angle be $x$. So, $$70^\circ + x + x = 180^\circ$$ $$70^\circ + 2x = 180^\circ$$ $$2x = 110^\circ$$ $$x = 55^\circ$$ Thus, $\angle ABD = 55^\circ$. 6. **Step 4: Find $\angle ABC$** $\angle ABC$ is composed of $\angle ABD$ and $\angle DBC$. We know $\angle ABD = 55^\circ$ from above. $\angle DBC$ is part of $\triangle BCD$, and since $\triangle BCD$ is equilateral, $\angle DBC = 60^\circ$. 7. **Step 5: Calculate $\angle ABC$** $$\angle ABC = \angle ABD + \angle DBC = 55^\circ + 60^\circ = 115^\circ$$ **Final answer:** $$\boxed{115^\circ}$$