1. **State the problem:** Given parallelogram ABCD with $m\angle DAC = 28^\circ$, $m\angle DCA = 25^\circ$, $AD = 18$, $AB = 10$, and $BE = 8$ on diagonal $BD$, find $m\angle ABC$ and length $BD$. 2. **Recall properties:** In parallelogram ABCD, opposite sides are equal and opposite angles are equal. Also, diagonal $BD$ splits the parallelogram into two triangles. 3. **Find $m\angle A$:** Since $m\angle DAC = 28^\circ$ and $m\angle DCA = 25^\circ$, triangle $ADC$ has angles $28^\circ$, $25^\circ$, and $m\angle ADC = 180^\circ - 28^\circ - 25^\circ = 127^\circ$. 4. **Find $BD$ using Law of Cosines in triangle $ADC$:** $$BD^2 = AD^2 + DC^2 - 2 \times AD \times DC \times \cos(m\angle ADC)$$ Since $DC = AB = 10$ (opposite sides equal), $$BD^2 = 18^2 + 10^2 - 2 \times 18 \times 10 \times \cos(127^\circ)$$ Calculate: $$BD^2 = 324 + 100 - 360 \times \cos(127^\circ)$$ Since $\cos(127^\circ) \approx -0.6018$, $$BD^2 = 424 + 360 \times 0.6018 = 424 + 216.65 = 640.65$$ So, $$BD = \sqrt{640.65} \approx 25.31$$ 5. **Find $m\angle ABC$:** In parallelogram, $m\angle ABC = m\angle ADC = 127^\circ$. 6. **Check $BE = 8$:** Given $BE=8$ on $BD \approx 25.31$, consistent with the problem. 7. **Final answers:** $$m\angle ABC = 127^\circ$$ $$BD \approx 25.31$$ Among options, closest is D: $m\angle ABC = 127$, $BD = 16$ but $BD$ calculated is about 25.31, so none exactly match $BD=16$. However, angle matches option D. **Answer:** $m\angle ABC = 127^\circ$, $BD \approx 25.31$ (closest to option D angle).