Answer:
△ AMC ≅ △ BMC ⇒ proved down
Step-by-step explanation:
Let us 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 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
In Δ ABC
∵ CM ⊥ AB
∴ m∠AMC = m∠BMC = 90°
∴ ∠1 ≅ ∠2
In the two triangles AMC and BMC
∵ ∠1 ≅ ∠2 ⇒ proved
∵ ∠3 ≅ ∠4 ⇒ given
∵ CM is a common side in the two triangles
∴ CM ≅ CM ⇒ common side
→ Two angles and the side joining them in the 1st triangle ≅ two angles
and the side joining them in the 2nd triangle, then use the 3rd case
of congruency above
∴ Δ AMC ≅ Δ BMC ⇒ by using ASA postulate
Answer:
36, 2×2×3×3
Step-by-step explanation:
8 free throws= 8×1=8
14 2 point baskets= 14×2=28
8+28=36
8:2
Or you could also say these
8 to 2
8/2
Answer:
Step-by-step explanation:
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