1. **Stating the problem:** We have triangle ABC with sides AB = BC and a point D on the extension of BA such that AB = AD. We need to find the measure of angle BCD. 2. **Understanding the setup:** Since AB = BC, triangle ABC is isosceles with AB = BC. Point D is on line BA extended beyond A, and AD = AB. 3. **Assign variables:** Let the length of AB = BC = x. Because AD = AB, we also have AD = x. Since D lies beyond A on line BA, BD = BA + AD = x + x = 2x. 4. **Coordinates approach (for clarity):** Place B at origin (0,0). Since AB = x and BA is along the x-axis, place A at (x,0). Since triangle ABC is isosceles with AB=BC, point C lies somewhere above the x-axis such that BC = x. 5. **Find coordinates of C:** Because BC = x and B at (0,0), C lies on a circle centered at (0,0) with radius x. Also, AB = x is along the x-axis from B(0,0) to A(x,0). Since triangle is isosceles with AB = BC, C lies at (x/2, h) where h is height. 6. **Calculate height h:** Using the distance formula BC = x, $$\sqrt{\left(x/2 - 0\right)^2 + (h-0)^2} = x$$ $$\Rightarrow \sqrt{(x^2/4) + h^2} = x$$ $$\Rightarrow (x^2/4) + h^2 = x^2$$ $$\Rightarrow h^2 = x^2 - x^2/4 = (3x^2)/4$$ $$\Rightarrow h = \frac{\sqrt{3}x}{2}$$. Thus, C = \left( \frac{x}{2}, \frac{\sqrt{3}x}{2} \right). 7. **Coordinates of D:** Since D lies on extension of BA beyond A, and BA is on x-axis from B(0,0) to A(x,0), D is at (2x,0). 8. **Vector approach to find angle BCD:** We want angle BCD at point C formed by points B and D. Vectors: $$\vec{CB} = \vec{B} - \vec{C} = (0 - \frac{x}{2}, 0 - \frac{\sqrt{3}x}{2}) = \left(-\frac{x}{2}, -\frac{\sqrt{3}x}{2}\right)$$ $$\vec{CD} = \vec{D} - \vec{C} = (2x - \frac{x}{2}, 0 - \frac{\sqrt{3}x}{2}) = \left(\frac{3x}{2}, -\frac{\sqrt{3}x}{2}\right)$$ 9. **Calculate dot product:** $$\vec{CB} \cdot \vec{CD} = \left(-\frac{x}{2}\right)\left(\frac{3x}{2}\right) + \left(-\frac{\sqrt{3}x}{2}\right)\left(-\frac{\sqrt{3}x}{2}\right) = -\frac{3x^2}{4} + \frac{3x^2}{4} = 0$$ 10. **Calculate magnitudes:** $$|\vec{CB}| = \sqrt{\left(-\frac{x}{2}\right)^2 + \left(-\frac{\sqrt{3}x}{2}\right)^2} = \sqrt{\frac{x^2}{4} + \frac{3x^2}{4}} = \sqrt{x^2} = x$$ $$|\vec{CD}| = \sqrt{\left(\frac{3x}{2}\right)^2 + \left(-\frac{\sqrt{3}x}{2}\right)^2} = \sqrt{\frac{9x^2}{4} + \frac{3x^2}{4}} = \sqrt{3x^2} = x\sqrt{3}$$ 11. **Angle between vectors:** $$\cos(\angle BCD) = \frac{\vec{CB} \cdot \vec{CD}}{|\vec{CB}| |\vec{CD}|} = \frac{0}{x \cdot x\sqrt{3}} = 0$$ 12. **Conclusion:** Since the cosine of the angle is zero, $$\angle BCD = 90^{\circ}$$. **Final answer:** The measure of angle BCD is $90^{\circ}$.