Answer:
Δ MNO ≅ ΔXYZ ⇒ proved down
Step-by-step explanation:
Cases of congruency
- 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 1st Δ ≅ 2 angles and one side in the 2nd Δ
- HL ⇒ hypotenuse and leg of the 1st right Δ ≅ hypotenuse and leg of the 2nd right Δ
Let us solve the question using the 4th rule above
∵ YZ = YO + OZ
∵ NO = NZ + ZO
∵ OZ = ZO ⇒ from the figure
∵ YO ≅ NZ ⇒ given
∴ YZ ≅ NO
In Δs MNO and XYZ
∵ ∠M ≅ ∠X ⇒ given
∵ ∠N ≅ ∠Y ⇒ given
∵ NO ≅ YZ ⇒ proved
→ By using the AAS postulate of congruency.
∴ Δ MNO ≅ ΔXYZ
