Question 1179586: https://ibb.co/jgpq3pX (Please copy paste the link to view the figure)
Given: AE = DE; EB = EC
Prove: a. Triangle AEB = Triangle DEC
b. Triangle ABC = Triangle DCB
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Perfect, let's break down this geometric proof.
**Given:**
* AE = DE
* EB = EC
**Prove:**
* a. △AEB ≅ △DEC
* b. △ABC ≅ △DCB
**Proof:**
**a. △AEB ≅ △DEC**
| Statement | Reason |
|---|---|
| 1. AE = DE | Given |
| 2. EB = EC | Given |
| 3. ∠AEB ≅ ∠DEC | Vertical angles are congruent |
| 4. △AEB ≅ △DEC | Side-Angle-Side (SAS) Congruence Theorem (AE = DE, ∠AEB ≅ ∠DEC, EB = EC) |
**b. △ABC ≅ △DCB**
To prove this, we need to establish that AB = DC, AC = DB and BC is a shared side.
| Statement | Reason |
|---|---|
| 1. AE = DE | Given |
| 2. EB = EC | Given |
| 3. △AEB ≅ △DEC | Proven in part a |
| 4. AB = DC | Corresponding Parts of Congruent Triangles are Congruent (CPCTC) |
| 5. AC = AE + EC | Segment Addition Postulate |
| 6. DB = DE + EB | Segment Addition Postulate |
| 7. AC = DE + EB | Substitution Property (substitute AE with DE and EC with EB) |
| 8. AC = DB | Transitive Property of Equality (since AC = DE + EB and DB = DE + EB) |
| 9. BC = BC | Reflexive Property of Equality (any segment is congruent to itself) |
| 10. △ABC ≅ △DCB | Side-Side-Side (SSS) Congruence Theorem (AB = DC, AC = DB, BC = BC) |
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