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
