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)
