1. **State the problem:** We have a quadrilateral BACD with diagonals AD and BC drawn, dividing it into four triangles. Given that AB = BC = BD and BD is parallel to AC, and angles at B and C are 40° and 55° respectively, we need to find the angle at D, labeled as $x$. 2. **Analyze given information:** - Since AB = BC = BD, triangle BCD is isosceles with BC = BD. - BD || AC implies that angles formed by transversal lines have relationships. 3. **Use the parallel lines property:** Since BD || AC, angle ABD equals angle BAC (alternate interior angles), and angle BDC equals angle DCA. 4. **Calculate angle at B:** Given angle at B is 40°. 5. **Calculate angle at C:** Given angle at C is 55°. 6. **Use triangle properties:** In triangle BCD, since BC = BD, angles opposite these sides are equal. So angle BCD = angle BDC = $x$. 7. **Sum of angles in triangle BCD:** $$40° + x + x = 180°$$ $$2x = 180° - 40° = 140°$$ $$x = 70°$$ 8. **Final answer:** The angle at vertex D, $x$, is $70°$.