Read the proof. given: aeec; bddc prove: △aec ~ △bdc statement reason 1. aeec; bddc 1. given 2. ∠aec is a rt. ∠; ∠bdc is a rt. ∠ 2. definition of perpendicular 3. ∠aec ≅ ∠bdc 3. all right angles are congruent 4. ? 4. reflexive property 5. △aec ~ △bdc 5. aa similarity theorem what is the missing statement in step 4? ∠ace ≅ ∠bcd ∠eab ≅ ∠dbc ∠eac ≅ ∠eac ∠cbd ≅ ∠dbc

The missing Statement is

4. ∠EAC ≅ ∠EAC                               4. reflexive property

Step-by-step explanation:

Given:

AEEC; BDDC

To Prove:

△AEC ~ △BDC

Proof:

In △AEC  and △BDC

Statement                                              Reason

1. AE⊥EC;BD⊥DC                                  1. given

2. ∠AEC is a rt. ∠; ∠BDC is a rt. ∠    2. definition of perpendicular

3. ∠AEC ≅ ∠BDC                              3. all right angles are congruent

4. ∠EAC ≅ ∠EAC                               4. reflexive property

5. △AEC ~ △BDC                               5. AA similarity theorem

true? is that what you're asking or do you mean something else?