It is known that ad is the middle line of △ ABC, AE is the middle line of △ abd, ab = DC, ∠ bad = ∠ BDA I'm going to hand it in in the evening. Can anyone answer this question? It's not easy to draw pictures. You can draw by yourself. I only learned congruent triangles and axisymmetry. Please use my knowledge to solve it

It is known that ad is the middle line of △ ABC, AE is the middle line of △ abd, ab = DC, ∠ bad = ∠ BDA I'm going to hand it in in the evening. Can anyone answer this question? It's not easy to draw pictures. You can draw by yourself. I only learned congruent triangles and axisymmetry. Please use my knowledge to solve it

Take the midpoint F of AC and connect DF. Since D is the midpoint of BC, DF is the median line of triangle ABC, and DF is parallel to AC,
Then ∠ bad = ∠ FDA (equal internal stagger angle)
∠ B = ∠ CDF (same position angle)
Known ∠ bad = ∠ BDA
Introduce ∠ FDA = ∠ BDA
Because BC / AB = AB / be = 2 / 1, and ∠ CBA = ∠ Abe
The introduction of △ ABC is similar to △ EBA
Introduce ∠ C = ∠ BAE
In △ Abe and △ CDF
∠BAE=∠C
∠B=∠CDF
AC=CD
Deduce that △ Abe is equal to △ CDF (corner, side, corner)
Be = DF
Known be = de
Then de = DF
In △ ade and △ ADF
DE=DF
AD=AD
∠FDA=∠BDA
Deduce that △ ade is equal to △ ADF (edge)
Launch AE = AF
F is the midpoint of AC
So AC = 2ae