Han wrote a proof that triangle BCD is congruent to triangle DAB. Han's proof is incomplete. Which step is missing some information? DC∥AB
Step 1: Line AB is parallel to line DC and cut by transversal DB. So angles CDB and ABD are alternate interior angles and must be congruent.
Step 2: Side DB is congruent to side BD because they're the same segment.
Step 3: Angle A is congruent to angle C because they're both right angles.
Step 4: By the Angle-Side-Angle Triangle Congruence Theorem, triangle BCD is congruent to triangle DAB .
and WHY?

Alternate interior angles are angles formed at the inner corners of intersection on opposite sides of transversal when two parallel lines are intersected by a transversal.

Since Line AB is parallel to line DC and cut by transversal DB. So angles CDB and ABD are alternate interior angles and must be congruent.

Step 2:

DB ≅ BD (reflexive property of congruence)

Step 3:

∠A = ∠C = 90° (right angles, given). But to prove that both triangles are congruent, the correct step 3 is:

Since, ∠A = ∠C = 90° and angles CDB and ABD are congruent. This means that the third angles in the both triangles are also congruent. Hence, angle ADB and CBD are must be congruent.

Step 4:

BD = DB, angles CDB and ABD are congruent, angle ADB and CBD are congruent.

Therefore, By the Angle-Side-Angle Triangle Congruence Theorem, triangle BCD is congruent to triangle DAB .