1. **Problem Statement:** Find the value of $y$ in the triangle configuration where $\triangle ABC$ and $\triangle DBC$ are isosceles with $AB=AC$ and $DB=DC$, $E$ is on $BC$ such that $BE=EC$, $ADE$ is a straight line, $\angle BAC=46^\circ$, and $\angle EDC=37^\circ$. We need to find $\angle ABD$. 2. **Understanding the problem:** Since $AB=AC$, $\triangle ABC$ is isosceles with $\angle ABC=\angle ACB$. Similarly, $DB=DC$ means $\triangle DBC$ is isosceles with $\angle DBC=\angle DCB$. 3. **Step 1: Calculate angles in $\triangle ABC$:** Since $\angle BAC=46^\circ$, the base angles are equal: $$\angle ABC = \angle ACB = \frac{180^\circ - 46^\circ}{2} = \frac{134^\circ}{2} = 67^\circ.$$ 4. **Step 2: Analyze $\triangle DBC$:** Since $DB=DC$, $\angle DBC = \angle DCB$. Let each be $x$. 5. **Step 3: Use $E$ midpoint property:** $E$ is midpoint of $BC$, so $BE=EC$. 6. **Step 4: Use $\angle EDC=37^\circ$ and $ADE$ straight line:** Since $ADE$ is straight, $\angle ADC = 180^\circ - \angle EDC = 180^\circ - 37^\circ = 143^\circ$. 7. **Step 5: In $\triangle DBC$, sum of angles:** $$\angle DBC + \angle DCB + \angle BDC = 180^\circ$$ Since $\angle DBC = \angle DCB = x$ and $\angle BDC = 143^\circ$, $$2x + 143^\circ = 180^\circ \implies 2x = 37^\circ \implies x = 18.5^\circ.$$ 8. **Step 6: Find $\angle ABD$:** $\angle ABD$ is part of $\angle ABC$ which is $67^\circ$. Since $D$ lies on $BC$, and $BD=DC$, $\angle ABD = x = 18.5^\circ$. **Final answer:** $$\boxed{\angle ABD = 18.5^\circ}$$