Question

Given: EB = EC, AAED is equilateral and equiangular. Prove: AACD = ADBA

Answer 1

Statement Reason

1. Line EB is congruent to line EC 1. Given
2. Triangle AED is equilateral 2. Given
3. Line BE + Line ED = BD 3.Segment Addition Postulate
4. Line CE + Line EA = CA 4.Segment Add. Postulate
5. BD is congruent to CA 5. Transitive Property?
6. AD is congruent to AD 6. Reflexive Prop.
7. Angle EAD is congruent to Angle EDA 7. Def. of Equilateral Triangle?
8. Tri. ACD is congruent to Tri. DBA 8. SAS congruency

Answer 2

we can conclude that ΔACD ≅ ΔDBA.

To prove that ΔACD ≅ ΔDBA, we will use the Side-Side-Side (SSS) Congruence Postulate.

This postulate states that if two triangles have three corresponding sides that are congruent, then the triangles are congruent.

Given:

EB = EC

ΔAED is equilateral and equiangular

From statement 2, we can deduce the following:

3. AE = ED = AD (since ΔAED is equilateral)

4. ∠AED = ∠EAD = ∠ADE = 60° (since ΔAED is equiangular)

From statement 1, we have:

5. AE = EC

Now, we can apply the SSS Congruence Postulate:

Since we have three pairs of corresponding sides that are congruent (AE = EC, EC = BA, and AE = BA), we can conclude that ΔACD ≅ ΔDBA.