1pts
given: c is midpoint of bd
ab 1 bd
bd i de
prove: aabc = aedc
what will be the final reason" in this proof?
aas
asa
osas
h-l

ASA

Step-by-step explanation:

Given:

Two triangles ABC and EDC such that:

AB ⊥ BD and BD ⊥ DE

C is the midpoint of BD.

The two triangles are drawn below.

Since, AB ⊥ BD and BD ⊥ DE

Therefore, the two triangles are right angled triangle. The triangle ABC is right angled at vertex B. The triangle EDC is right angled at vertex D.

Since, point C is the midpoint of the line segment BD.

Therefore, C divides the line segment BD into two equal parts.

So, segment BC ≅ segment CD (Midpoint theorem)

Now, consider the triangles ABC and EDC.

Statements                                             Reason

1. ∠ABC ≅ ∠CDE                         Right angles are congruent to each other

2. BC ≅ CD                                   Midpoint theorem. C is midpoint of BD

3. ∠ACB ≅ ∠ECD                         Vertically opposite angles are congruent

Therefore, the two triangles are congruent by ASA postulate.

So, the second option is correct.

this is false ^.^ lol