QUESTION IMAGE
Question
tu ≅ tv. complete the proof that δtuv ≅ δtvu.
triangle with vertices v, u, t
| statement | reason |
|---|---|
| 2. ∠t ≅ ∠t | |
| 3. δtuv ≅ δtvu |
Step1: Identify Given Information
The problem states \(\overline{TU} \cong \overline{TV}\), so the reason for statement 1 is "Given".
Step2: Identify Reflexive Property
\(\angle T\) is common to both \(\triangle TUV\) and \(\triangle TVU\), so \(\angle T \cong \angle T\) by the "Reflexive Property of Congruence" (an angle is congruent to itself).
Step3: Identify Congruence Criterion
We have two sides? Wait, no, we have a side (\(\overline{TU} \cong \overline{TV}\)), an included angle? Wait, no, \(\overline{TU}\) and \(\overline{TV}\) with common angle \(\angle T\), and also \(\overline{TV}\) and \(\overline{TU}\) are the same as \(\overline{UT}\) and \(\overline{VT}\)? Wait, actually, \(\triangle TUV\) and \(\triangle TVU\) share side \(\overline{TU}\) and \(\overline{TV}\) (given congruent), and angle \(\angle T\) (common). Wait, the triangles are \(\triangle TUV\) and \(\triangle TVU\), so the sides: \(TU = TV\) (given), \(T\) is common, and \(UV\) and \(VU\) are the same side (reflexive). Wait, no, actually, the congruence is by SAS? Wait, \(TU = TV\) (given), \(\angle T = \angle T\) (reflexive), and \(TV = TU\) (same as given), so actually, the triangles are congruent by SAS (Side - Angle - Side) because we have two sides and the included angle. Wait, the third side: \(\overline{UV}\) and \(\overline{VU}\) are the same, so actually, the reason for statement 3 is "SAS (Side - Angle - Side) Congruence Postulate" because we have \(TU = TV\) (side), \(\angle T=\angle T\) (angle), and \(TV = TU\) (side) – wait, no, the triangles are \(\triangle TUV\) and \(\triangle TVU\), so the correspondence is \(T \to T\), \(U \to V\), \(V \to U\). So \(TU = TV\) (given), \(\angle T=\angle T\) (reflexive), and \(UV = VU\) (reflexive, since it's the same side). Wait, no, that would be SSS, but we have a side and an angle. Wait, no, the correct correspondence: in \(\triangle TUV\) and \(\triangle TVU\), the sides are \(TU\) and \(TV\) (given congruent), \(T\) is common, and \(UV\) and \(VU\) are congruent (reflexive). But actually, the SAS postulate: if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. Here, \(TU = TV\) (side), \(\angle T=\angle T\) (angle), and \(TV = TU\) (side) – wait, no, the included angle for \(TU\) and \(TV\) is \(\angle T\), so yes, SAS: \(TU \cong TV\), \(\angle T \cong \angle T\), and \(TV \cong TU\) (same as \(TU \cong TV\)), so the triangles are congruent by SAS.
So to summarize:
- \(\overline{TU} \cong \overline{TV}\) - Given
- \(\angle T \cong \angle T\) - Reflexive Property of Congruence
- \(\triangle TUV \cong \triangle TVU\) - SAS (Side - Angle - Side) Congruence Postulate
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- Reason: Given
- Reason: Reflexive Property of Congruence
- Reason: SAS (Side - Angle - Side) Congruence Postulate