Congruent Triangle Combination

Answer:

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      [tex]{}[/tex]        given

Side BC ≅ Side BC      [tex]{}[/tex]            reflexive property

∠DBC ≅ ∠ACB                [tex]{}[/tex]        given

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

2) Side AC ≅ Side BD      [tex]{}[/tex]        given

Side DC ≅ Side BA      [tex]{}[/tex]            given

Side BC ≅ Side BC      [tex]{}[/tex]            reflexive property

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

3) ∠BAC ≅ ∠DEB                [tex]{}[/tex]        given

Side BC ≅ Side BD      [tex]{}[/tex]               given

∠ABC ≅ ∠DBE                [tex]{}[/tex]            vertically opposite angles

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

4) ΔXYZ ≅ ΔABC                [tex]{}[/tex]        given

m∠A = 44°, and m∠C = 83°

m∠B = 180 - (44 + 83) = 53°                [tex]{}[/tex]   Sum of angles in a triangle

m∠Y = m∠B = 53°                [tex]{}[/tex]   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      [tex]{}[/tex]           given

Side AC ≅ Side DC      [tex]{}[/tex]              given

∠ACB ≅ ∠DCE                [tex]{}[/tex]            vertically opposite angles

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

7) ∠BAC ≅ ∠DAC                [tex]{}[/tex]        given

∠ADC ≅ ∠ABC                [tex]{}[/tex]            given

Side AC ≅ Side AC      [tex]{}[/tex]               reflexive property

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

8) Side AD ≅ Side BC      [tex]{}[/tex]           given

Side AC ≅ Side AC      [tex]{}[/tex]               reflexive property

Side AB ≅ Side DC      [tex]{}[/tex]                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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