Given: ab parallel to dc prove: abe ~ dce

The reasons would be:

alternate interior angles
vertical angles
AAA (all angles are congruent)

Since we know that the two line segments are parallel, this means that BC is a transversal.  Using that knowledge with the position of ABE and DCE, we see these are alternate interior angles.

AEB and DEC are vertical angles, since they share only a vertex.

This gives us 3 congruent angles, so the triangles must be similar.

The 3 missing statements are listed below with explanations.

2) Alternate interior angles are congruent. Since the lines are parallel these angles are congruent.

3) Vertical angles are congruent. Intersecting angles across from each other are congruent.

4) AA Similarity Theorem. If 2 angles of a triangle are congruent, then the triangles are similar.

Shania is correct.

Step-by-step explanation:

For a right angled triangle, given two sides and an angle, we can use the sine rule or the various trigonometry identities to solve the triangle;

a/sin A = b/sinB

and

sin (angle) = opposite side/hypotenuse

cos (angle) = adjacent side/hypotenuse

tan (angle) = opposite/ adjacent

assuming a is the opposite side of angle A and b is also opposite of angle B

Otherwise, the only way to solve a triangle given two sides only is by the use of the Pythagoras theorem.

Hence, Shania is correct.