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Question
25 given: $overline{wy}=overline{xz}$ prove: $overline{wx}=overline{yz}$ statements reasons 1. $overline{wy}=overline{xz}$ 1. 2. $wy = xz$ 2. 3. $wx+xz = wz; wy + yz = wz$ 3. 4. $xz+yz = wz$ 4. 5. $wx+xz = xz + yz$ 5. 6. $wx = yz$ 6. 7. $overline{wy}=overline{xz}$ 7. 26 given: $d$ is the mid - point of prove: $de=\frac{1}{2}ce$ statements reasons 1. $d$ is the mid - point of 1. 2. $cd = de$ 2. 3. $cd + de = de+de$ 3. 4. $cd + de = 2de$ 4. 5. $cd + de = ce$ 5. 6. $2de = ce$ 6. 7. $de=\frac{1}{2}ce$ 7.
Step1: Given information
Given $\overline{WY}=\overline{XZ}$
Step2: Definition of congruent segments
If $\overline{WY}\cong\overline{XZ}$, then $WY = XZ$ (Definition of congruent - segments: congruent segments have equal lengths)
Step3: Segment - addition postulate
$WX+XZ = WZ$ and $WY + YZ=WZ$ (Segment - addition postulate: if a point lies on a line segment, the sum of the lengths of the two sub - segments equals the length of the whole segment)
Step4: Substitution
Since $WY = XZ$, we can substitute $WY$ with $XZ$ in $WY + YZ=WZ$, getting $XZ + YZ=WZ$
Step5: Transitive property of equality
Since $WX+XZ = WZ$ and $XZ + YZ=WZ$, then $WX+XZ=XZ + YZ$ (Transitive property: if $a = c$ and $b = c$, then $a = b$)
Step6: Subtraction property of equality
Subtract $XZ$ from both sides of $WX+XZ=XZ + YZ$, we get $WX=YZ$ (Subtraction property: if $a + c=b + c$, then $a = b$)
Step7: Definition of congruent segments
Since $WX = YZ$, then $\overline{WX}\cong\overline{YZ}$ (Definition of congruent segments: segments with equal lengths are congruent)
Step1: Given information
Given that $D$ is the mid - point of $\overline{CE}$
Step2: Definition of mid - point
If $D$ is the mid - point of $\overline{CE}$, then $CD = DE$ (Definition of mid - point: a mid - point of a line segment divides the segment into two equal parts)
Step3: Substitution
Since $CD = DE$, we can substitute $CD$ with $DE$ in $CD+DE$, getting $CD + DE=DE + DE$
Step4: Simplification
$DE+DE = 2DE$ (Combining like terms)
Step5: Segment - addition postulate
$CD + DE=CE$ (Segment - addition postulate: if a point lies on a line segment, the sum of the lengths of the two sub - segments equals the length of the whole segment)
Step6: Substitution
Substitute $CD + DE$ with $CE$ and $CD + DE$ with $2DE$, we get $2DE=CE$
Step7: Division property of equality
Divide both sides of $2DE = CE$ by 2, we get $DE=\frac{1}{2}CE$ (Division property: if $a = b$, then $\frac{a}{c}=\frac{b}{c}$, $c
eq0$)
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- Given
- Definition of congruent segments
- Segment - addition postulate
- Substitution
- Transitive property of equality
- Subtraction property of equality
- Definition of congruent segments
For the second problem: