This problem can be simply solved by just looking at it if you know that
- In a right-angled triangle, if a perpendicular is drawn from the right angle to the hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle (and are also similar to each other.)
This can be proved as follows,

In the triangle ABC, (angle BAD + angle BCD) = 90° ................ (1)
In the triangle ABD, (angle BAD + angle ABD) = 90° ................ (2)
In the triangle CBD, (angle CBD + angle BCD) = 90° ................ (3)
From (1) and (2) ---> angle BCD = angle ABD
From (1) and (3) ---> angle BAD = angle CBD
Hence, all the three triangles are similar triangles.
Now if we go on drawing perpendiculars to hypotenuse of the children triangles, all the triangles we will get will be similar to original triangle and all the other triangles.
Hence, in our question triangle BDE will be similar to triangle BAD. (This can be proved separately as others did but we don't need to do it as we know it has to be). Now, from the question stem we know the length of the side which is common to these two triangles. Hence, if we know the length of another side of triangle BDE, we can determine the length of AB.
Hence, both statements are individually sufficient.
The correct answer is D.