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)
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)