1. **Problem Statement:** ABCD is a rhombus where the altitude from vertex D to side AB bisects AB. We need to find the angles of the rhombus. 2. **Understanding the problem:** A rhombus has all sides equal and opposite angles equal. The altitude from D to AB is perpendicular to AB and bisects AB, meaning it hits AB at its midpoint. 3. **Key properties and formula:** - Let the side length be $s$. - Since altitude bisects AB, let midpoint be $M$. - Altitude $DM$ is perpendicular to $AB$. - In rhombus, diagonals bisect each other at right angles. 4. **Set up coordinate system:** - Place $A$ at $(0,0)$ and $B$ at $(2a,0)$ so that $M$ is at $(a,0)$. - Since $DM$ is altitude, $D$ lies on the line perpendicular to $AB$ at $M$, so $D$ is at $(a,h)$. 5. **Using side length equality:** - $AB = 2a$. - $AD = s = \sqrt{(a-0)^2 + (h-0)^2} = \sqrt{a^2 + h^2}$. - $BD = s = \sqrt{(2a - a)^2 + (0 - h)^2} = \sqrt{a^2 + h^2}$. 6. **Since $AB$ is side length $s$, we have:** $$s = 2a$$ 7. **From above, $AD = s = \sqrt{a^2 + h^2}$, so:** $$\sqrt{a^2 + h^2} = 2a$$ 8. **Square both sides:** $$a^2 + h^2 = 4a^2$$ $$h^2 = 3a^2$$ $$h = a\sqrt{3}$$ 9. **Find angles:** - Angle at $A$ is between $AB$ and $AD$. - Vector $AB = (2a,0)$. - Vector $AD = (a,h) = (a, a\sqrt{3})$. 10. **Calculate angle $\theta$ between $AB$ and $AD$ using dot product:** $$\cos \theta = \frac{AB \cdot AD}{|AB||AD|} = \frac{2a \times a + 0 \times a\sqrt{3}}{2a \times 2a} = \frac{2a^2}{4a^2} = \frac{1}{2}$$ 11. **Therefore:** $$\theta = \cos^{-1}(\frac{1}{2}) = 60^\circ$$ 12. **Since opposite angles are equal and adjacent angles are supplementary:** - Angles of rhombus are $60^\circ$ and $120^\circ$. **Final answer:** The angles of the rhombus are $60^\circ$ and $120^\circ$.