Triangle BAC was dilated from triangle BDE at a scale factor of 2. What proportion proves that \( \cos \angle D=\cos \angle A \) ?

Step 1 ：\( \triangle BDE \sim \triangle BAC \) , scale factor = 2

Step 2 ：\( \frac{DE}{AC} = \frac{BD}{AB} = \frac{BE}{BC} = \frac{1}{2} \)

Step 3 ：\( \cos \angle D = \frac{BD}{DE} , \cos \angle A = \frac{AB}{AC} \)

Step 4 ：\( \cos \angle D = \cos \angle A \) since \( \frac{BD}{DE} = \frac{AB}{AC} \)