Answer:
ASA
ΔFGH ≅ ΔIHG ⇒ answer B
Step-by-step explanation:
* Lets revise the cases of congruence
- 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 prove the two triangles FGH and IHG are congruent by on of
the cases above
∵ FG // HI and GH is transversal
∴ m∠FGH = m∠IHG ⇒ alternate angles
- In the two triangles FGH and IHG
∵ m∠FHG = m∠IGH ⇒ given
∵ m∠FGH = m∠IHG ⇒ proved
∵ GH = HG ⇒ common side
∴ ΔFGH ≅ ΔIHG ⇒ ASA
* ASA
ΔFGH ≅ ΔIHG
