What are you confused about? It is impossible to give you any help or suggestions without knowing what you can do. You are given a rectangle, ACDE, with B the midpoint of side AC. Lines BD and BE are drawn. You are asked to prove that triangles BAD and BCE are congruent.
Okay, what do you know about congruent triangles? Do you know "Side-Side-Side", "Side-Angle-Side", and "Angle-Side-Angle". Since any two right triangles have the right angles congruent, as well as having the third side determined by the other two, there are special congruence theorems for right angles.
i have so far
1.angle BED=angleBDE given
2.Triangle BDE is isosceles angles BED and BDE are equal
3.Side BE=Side BD triangle BDE is isosceles
4.triangle EAB=triangle DCB triangles are congruent
5.BA=BC, BE=BD each triangle is right angled
Actually ACDE is not mentioned to be a rectangle in the original problem. It can be proved that it is, though I wouldn't use that fact at the start of the proof. (I know you didn't, I'm just being fussy about the matter. )
@Plato: I learned two column proofs when I was in HS about 25 years ago. Just for the record.
-Dan
Actually it can. However it is not needed to solve the problem. (The proof is also a bit longer than I had thought.)
Quickly speaking we can show that triangles EAB and DCB are congruent. Doing some angle work we can show that angles BED and ABE are equal. We now have two lines (AC and ED) cut by a transversal (BE) and alternate interior angles are the same so AD and ED are parallel. From here take AE to be a transversal cutting across two parallel lines (AC and ED again), where BAE is a right angle, and thus so is DEA. etc. So we have a 4 sided figure that has right angles on each corner. Thus ACDE is a rectangle.
-Dan