Respuesta :

Answer:

Δ TUW ≅ ΔVWU ⇒ by AAS case

Step-by-step explanation:

* Lets revise the cases of congruent for triangles

- SSS  ⇒ 3 sides in the 1st Δ ≅ 3 sides in the 2nd Δ

- SAS ⇒ 2 sides and including angle in the 1st Δ ≅ 2 sides and  

 including angle in the 2nd Δ  

- ASA ⇒ 2 angles and the side whose joining them in the 1st Δ  

 ≅ 2 angles and the side whose joining them in the 2nd Δ  

- AAS ⇒ 2 angles and one side in the first triangle ≅ 2 angles

 and one side in the 2ndΔ

- HL ⇒ hypotenuse leg of the first right angle triangle ≅ hypotenuse

 leg of the 2nd right angle Δ

* Lets solve the problem

- There are two triangles TUW and VWU

- ∠T and ∠V are right angles

- LINE TW is parallel to line VU

∵ TW // VU and UW is a transversal

∴ m∠VUW = m∠TWU ⇒ alternate angles (Z shape)

- Now we have in the two triangles two pairs of angle equal each

 other and one common side, so we can use the case AAS

- In Δ TUW and ΔVWU

∵ m∠T = m∠V ⇒ given (right angles)

∵ m∠TWU = m∠VUW ⇒ proved

∵ UW = WU ⇒ (common side in the 2 Δ)

∴ Δ TUW ≅ ΔVWU ⇒ by AAS case

Answer:

Step-by-step explanation:

Given ∠T and ∠V are right angles.

TW ║ UV

To prove ⇒ ΔTUW ≅ ΔVWU

Proof ⇒ In ΔTUW and ΔVUW,

∠T ≅ ∠ V ≅ 90° (given)

Side UW ≅ UW ( Common in both the triangles )

TW ║ UV

and UW is a transverse.

So ∠TWU ≅ ∠WUV [alternate interior angles]

Since Angle = Angle = side are equal

Therefore, ΔTUW ≅ ΔVWU

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