Question

given: (overline{wy}congoverline{xz}) prove: (overline{wx}congoverline{yz}) 1. (overline{wy}congoverline{xz}) given 2. (wy = xz) 3. (wx+xz = wz) (wy+yz = wz) 4. (xz+yz = wz) 5. (wx+xz = xz+yz) 6. (wx = yz) 7. (overline{wx}congoverline{yz}) # segment addition property # def of congruence # subtraction property # substitution # def of congruence # transitive property

Explanation:

Step1: <Def of Congruence>

$$\overline{WY}\cong\overline{XZ}\implies WY = XZ$$

Step2: <Segment Addition Property>

$$WX+XY = WY,\quad XY + YZ=XZ$$

Step3: <Substitution>

Since \(WY = XZ\), substitute \(WY\) with \(XZ\) in \(WX + XY=WY\) to get \(WX+XY=XZ\). Then, since \(XY + YZ = XZ\), we have \(WX+XY=XY + YZ\)

Step4: <Subtraction Property>

Subtract \(XY\) from both sides of \(WX+XY=XY + YZ\). So \(WX+XY - XY=XY + YZ - XY\)

Step5: <Simplify>

\(WX=YZ\)

Step6: <Def of Congruence>

If \(WX = YZ\), then \(\overline{WX}\cong\overline{YZ}\)

Answer:

1. \(\text{Def of Congruence}\)；3. \(\text{Segment Addition Property}\)；4. \(\text{Segment Addition Property}\)；5. \(\text{Substitution}\)；6. \(\text{Subtraction Property}\)；7. \(\text{Def of Congruence}\)