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 ≅ ∠ace
∠eab ≅ ∠dbc
∠eac ≅ ∠eac
∠cbd ≅ ∠dbc

Okay so It is given that ∠ABR and ∠ACR are right angles, AB ≅ BC and BC ≅ AC Since they contain right angles, △ABR and △ACR are right triangles. The right triangles share hypotenuse AR, and reflexive property justifies that AR ≅ AR. Since AB ≅ BC and BC ≅ AC, the transitive property justifies AB ≅ AC. Now, the hypotenuse and leg of right △ABR is congruent to the hypotenuse and the leg of right △ACR, so △ABR ≅ △ACR by the HL congruence postulate. Therefore, ∠BAR ≅ ∠CAR by CPCTC, and bisects ∠BAC by the definition of bisector.