Question

3 given: x is the mid - point of wy and vz. prove: ∠xwy≅∠xyz. statements reasons 1. x is midpoint of wy and vz 1. given 2. vx≅xz 2. def of mid - point 3. wx≅xy 3. def of mid - point 4. ∠wxv≅∠zxy 4. vertical angle theorem 5. 5. 6. ∠xwy≅∠xyz 6. 4 given: jm bisects ∠kjl, ∠jmk≅∠jml. prove: jk≅jl. statements reasons 1. jm bisects ∠jkl 1. given 2. ∠jmk≅∠jml 2. given 3. jm≅jm 3. reflexive 4. 4. 5. 5. 6. jk≅jl 6. 5 given: bc∥ef, d is the mid - point of bf. prove: ed≅cd. statements reasons 1. bc∥ef 1. given 2. d is midpoint of bf 2. given 3. bd≅df 3. definition of midpoint 4. 4. 5. 5. 6. 6. 7.

Explanation:

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Step1: Given mid - point

Since \(X\) is the mid - point of \(\overline{WY}\) and \(\overline{VZ}\), by the definition of mid - point, we have \(\overline{VX}\cong\overline{XZ}\) and \(\overline{WX}\cong\overline{XY}\).

Step2: Vertical angles

\(\angle WXV\) and \(\angle ZXY\) are vertical angles. By the vertical - angle theorem, \(\angle WXV\cong\angle ZXY\).

Step3: SAS congruence

In \(\triangle XWV\) and \(\triangle XYZ\), we have \(\overline{WX}\cong\overline{XY}\), \(\angle WXV\cong\angle ZXY\), and \(\overline{VX}\cong\overline{XZ}\). So, by the Side - Angle - Side (SAS) congruence postulate, \(\triangle XWV\cong\triangle XYZ\).

Step4: Corresponding parts of congruent triangles

Since \(\triangle XWV\cong\triangle XYZ\), corresponding angles are congruent. So, \(\angle XWV\cong\angle XYZ\).

Step1: Given angle bisector

Since \(\overline{JM}\) bisects \(\angle KJL\), by the definition of an angle bisector, \(\angle KJM\cong\angle LJM\).

Step2: Given equal angles

We are given that \(\angle JMK\cong\angle JML\).

Step3: Reflexive property

\(\overline{JM}\cong\overline{JM}\) by the reflexive property of congruence.

Step4: ASA congruence

In \(\triangle JMK\) and \(\triangle JML\), we have \(\angle KJM\cong\angle LJM\), \(\overline{JM}\cong\overline{JM}\), and \(\angle JMK\cong\angle JML\). So, by the Angle - Side - Angle (ASA) congruence postulate, \(\triangle JMK\cong\triangle JML\).

Step5: Corresponding parts of congruent triangles

Since \(\triangle JMK\cong\triangle JML\), corresponding sides are congruent. So, \(\overline{JK}\cong\overline{JL}\).

Step1: Given parallel lines

Since \(\overline{BC}\parallel\overline{EF}\), by the alternate - interior angles theorem, \(\angle CBD\cong\angle EFD\).

Step2: Given mid - point

Since \(D\) is the mid - point of \(\overline{BF}\), by the definition of a mid - point, \(\overline{BD}\cong\overline{DF}\).

Step3: Vertical angles

\(\angle BDC\) and \(\angle FDE\) are vertical angles. By the vertical - angle theorem, \(\angle BDC\cong\angle FDE\).

Step4: ASA congruence

In \(\triangle BDC\) and \(\triangle FDE\), we have \(\angle CBD\cong\angle EFD\), \(\overline{BD}\cong\overline{DF}\), and \(\angle BDC\cong\angle FDE\). So, by the Angle - Side - Angle (ASA) congruence postulate, \(\triangle BDC\cong\triangle FDE\).

Step5: Corresponding parts of congruent triangles

Since \(\triangle BDC\cong\triangle FDE\), corresponding sides are congruent. So, \(\overline{ED}\cong\overline{CD}\).

Answer:

1. \(\triangle XWV\cong\triangle XYZ\); SAS congruence postulate
2. \(\angle XWV\cong\angle XYZ\); Corresponding parts of congruent triangles are congruent (CPCTC)

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