Question

complete the proofs below using the most appropriate method, sss or sas.

1. given: \\(\overline{ab} \cong \overline{ed}\\), \\(\overline{ab} \parallel \overline{de}\\), \\(c\\) is the midpoint of \\(\overline{ae}\\)

prove: \\(\triangle abc \cong \triangle edc\\)

statements | reasons

1. \\( yx \cong xz \\) |
2. \\( wx \\) bisects \\( \angle yxz \\) |
3. \\( \angle yxw \cong \angle zxw \\) |
4. \\( xw \cong xw \\) |
5. \\( \triangle wyx \cong \triangle wzx \\) |

given definition of midpoint reflexive property vertical angles definition of angle bisector
alternate interior angles alternate exterior angles corresponding angles sss sas

Explanation:

Step1: Analyze Statement 1

The statement \( YX \cong XZ \) is given in the problem's context (or from the problem's initial information), so the reason is "Given".

Step2: Analyze Statement 2

The statement "WX bisects \( \angle YXZ \)" is also part of the given information (or from the problem's setup), so the reason is "Given".

Step3: Analyze Statement 3

Since WX bisects \( \angle YXZ \), by the "Definition of Angle Bisector", we know that \( \angle YXW \cong \angle ZXW \).

Step4: Analyze Statement 4

The side \( XW \) is common to both \( \triangle WYX \) and \( \triangle WZX \), so by the "Reflexive Property", \( XW \cong XW \).

Step5: Analyze Statement 5

We have \( YX \cong XZ \), \( \angle YXW \cong \angle ZXW \), and \( XW \cong XW \). This satisfies the SAS (Side - Angle - Side) congruence criterion, so the reason for \( \triangle WYX \cong \triangle WZX \) is "SAS".

For the original problem (proving \( \triangle ABC \cong \triangle EDC \)):

Step1: Identify Given Information

We know that \( \overline{AB} \cong \overline{ED} \) (Given), \( \overline{AB} \parallel \overline{DE} \) (Given), and \( C \) is the midpoint of \( \overline{AE} \) (Given). Since \( C \) is the midpoint of \( \overline{AE} \), \( AC = EC \) (Definition of Midpoint).

Step2: Find the Included Angle

Because \( \overline{AB} \parallel \overline{DE} \), \( \angle ABC \cong \angle EDC \) (Alternate Interior Angles, as \( AB \parallel DE \) and \( BC \) and \( DC \) are transversals). Wait, actually, the vertical angles \( \angle ACB \) and \( \angle ECD \) are congruent (Vertical Angles).

Step3: Apply SAS

We have \( AB\cong ED \) (Given), \( \angle ACB\cong\angle ECD \) (Vertical Angles), and \( AC = EC \) (Definition of Midpoint). So by SAS (Side - Angle - Side) congruence criterion, \( \triangle ABC\cong\triangle EDC \).

But for the fill - in - the - blanks in the provided table (the \( \triangle WYX \) and \( \triangle WZX \) proof):

1. \( YX\cong XZ \): Reason - Given
2. \( WX \) bisects \( \angle YXZ \): Reason - Given
3. \( \angle YXW\cong\angle ZXW \): Reason - Definition of Angle Bisector
4. \( XW\cong XW \): Reason - Reflexive Property
5. \( \triangle WYX\cong\triangle WZX \): Reason - SAS

Answer:

1. Given
2. Given
3. Definition of Angle Bisector
4. Reflexive Property
5. SAS

(For the original \( \triangle ABC\cong\triangle EDC \) proof, the steps would be:

1. \( \overline{AB}\cong\overline{ED} \): Given
2. \( C \) is the midpoint of \( \overline{AE} \): Given, so \( AC = EC \): Definition of Midpoint
3. \( \overline{AB}\parallel\overline{DE} \): Given, so \( \angle ACB=\angle ECD \): Vertical Angles
4. \( \triangle ABC\cong\triangle EDC \): SAS)