1. **Stating the problem:** We are given angles $\angle BAD = (2a + 25)^\circ$ and $\angle BCD = (3a - 15)^\circ$. We need to find: a. The value of $a$ b. The measure of $\angle BAD$ c. The measure of $\angle CBA$ Also, diagonals $AC$ and $BD$ intersect at $E$ with $DE = 8$ cm and $AC = 13$ cm. We need to find: a. The length of $BD$ b. The length of $AE$ 2. **Finding $a$:** Since $\angle BAD$ and $\angle BCD$ are opposite angles in a parallelogram, they are equal. So, $$2a + 25 = 3a - 15$$ Rearranging, $$25 + 15 = 3a - 2a$$ $$40 = a$$ 3. **Finding $m \angle BAD$:** Substitute $a = 40$ into $2a + 25$: $$2(40) + 25 = 80 + 25 = 105^\circ$$ 4. **Finding $m \angle CBA$:** In a parallelogram, adjacent angles are supplementary, so $$m \angle BAD + m \angle CBA = 180^\circ$$ $$105 + m \angle CBA = 180$$ $$m \angle CBA = 180 - 105 = 75^\circ$$ 5. **Finding length of $BD$:** Diagonals of a parallelogram bisect each other, so $DE = EB = 8$ cm. Thus, $$BD = DE + EB = 8 + 8 = 16 \text{ cm}$$ 6. **Finding length of $AE$:** Since $E$ is midpoint of $AC$, $$AE = EC = \frac{AC}{2} = \frac{13}{2} = 6.5 \text{ cm}$$ **Properties applied:** - Opposite angles in a parallelogram are equal. - Adjacent angles in a parallelogram are supplementary. - Diagonals of a parallelogram bisect each other. **Final answers:** - $a = 40$ - $m \angle BAD = 105^\circ$ - $m \angle CBA = 75^\circ$ - $BD = 16$ cm - $AE = 6.5$ cm