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tw bisects ∠utv. complete the proof that △tvw ≅ △tuw. 1 tw bisects ∠utv…

Question

tw bisects ∠utv. complete the proof that △tvw ≅ △tuw.
1 tw bisects ∠utv
2 tu ≅ tv
3 ∠utw ≅ ∠vtw
4 tw ≅ tw
5 △tvw ≅ △tuw

Explanation:

Step1: Given information

The problem states that \(\overleftrightarrow{TW}\) bisects \(\angle UTV\), so the reason for statement 1 is "Given".

Step2: Given information

From the diagram, we can see that \(\overline{TU}\cong\overline{TV}\) (marked with the same tick - mark). So the reason for statement 2 is "Given".

Step3: Definition of angle bisector

Since \(\overleftrightarrow{TW}\) bisects \(\angle UTV\), by the definition of an angle bisector (\(\angle UTV=\angle UTVW + \angle VTVW\) and \(\angle UTVW=\angle VTVW\)), the reason for statement 3 is "Definition of angle bisector".

Step4: Reflexive property of congruence

For any segment \(AB\), \(AB\cong AB\). So for \(\overline{TW}\), \(\overline{TW}\cong\overline{TW}\) by the reflexive property of congruence. The reason for statement 4 is "Reflexive property of congruence".

Step5: SAS (Side - Angle - Side) congruence criterion

We have \(\overline{TU}\cong\overline{TV}\) (side), \(\angle UTW\cong\angle VTW\) (angle), and \(\overline{TW}\cong\overline{TW}\) (side). By the SAS (Side - Angle - Side) congruence criterion (\(\triangle ABC\cong\triangle DEF\) if \(AB = DE\), \(\angle B=\angle E\), and \(BC = EF\)), \(\triangle TVW\cong\triangle TUW\). The reason for statement 5 is "SAS (Side - Angle - Side) congruence criterion".

Answer:

  1. Given; 2. Given; 3. Definition of angle bisector; 4. Reflexive property of congruence; 5. SAS (Side - Angle - Side) congruence criterion.