Congruent Triangle Combination

1) ΔABC ≅ ΔDBC by Side-Angle-Side (SAS)

2) ΔABC ≅ ΔCDB by Side-Side-Side (SSS)

3) ΔABC ≅ ΔDBE by Angle-Angle-Side (AAS)

4) m∠Y = 53°

5) y = 3

6) ΔABC ≅ ΔEDC by Side-Angle-Side (SAS)

7) ΔABC ≅ ΔADC by Angle-Angle-Side (AAS)

8) ΔABC ≅ ΔADC by Side-Angle-Side (SSS)

Step-by-step explanation:

1) Side AC ≅ Side BD              given

Side BC ≅ Side BC                  reflexive property

∠DBC ≅ ∠ACB                        given

Therefore ΔABC ≅ ΔDBC by Side-Angle-Side (SAS)

2) Side AC ≅ Side BD              given

Side DC ≅ Side BA                  given

Side BC ≅ Side BC                  reflexive property

Therefore, ΔABC ≅ ΔCDB by Side-Side-Side (SSS)

3) ∠BAC ≅ ∠DEB                        given

Side BC ≅ Side BD                     given

∠ABC ≅ ∠DBE                            vertically opposite angles

Therefore, ΔABC ≅ ΔDBE by Angle-Angle-Side (AAS)

4) ΔXYZ ≅ ΔABC                        given

m∠A = 44°, and m∠C = 83°

m∠B = 180 - (44 + 83) = 53°                   Sum of angles in a triangle

m∠Y = m∠B = 53°                   Corresponding Parts of Congruent Triangles are Congruent (CPCTC)

5) Given that Triangle GHK is congruent to Triangle TZK

GH ≅ XT, KG ≅ KT, and HK ≅ ZK by the definition of congruency

Therefore, GH = 4 = XT = 3y - 2

4 = 3y - 2

3y - 2 = 4

3y = 4 + 2 = 6

y = 6/2 = 3

y = 3

6) Side BC ≅ Side EC                 given

Side AC ≅ Side DC                    given

∠ACB ≅ ∠DCE                            vertically opposite angles

Therefore, ΔABC ≅ ΔEDC by Side-Angle-Side (SAS)

7) ∠BAC ≅ ∠DAC                        given

∠ADC ≅ ∠ABC                            given

Side AC ≅ Side AC                     reflexive property

Therefore, ΔABC ≅ ΔADC by Angle-Angle-Side (AAS)

8) Side AD ≅ Side BC                 given

Side AC ≅ Side AC                     reflexive property

Side AB ≅ Side DC                      given that the included angle between the two legs side AD and side AC and side AC and side BC are both acute, the third sides side AB and side DC are equal

Therefore, ΔABC ≅ ΔADC by Side-Angle-Side (SSS)

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