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:
C
Step-by-step explanation:
y=mx+b
m(slope)=rise/run=1
b(y intercept)=2
y=x+2
The last one since the first one is up 13 and the last one is down 10
First, subtract bh from both sides and cancel. Then, divide both sides by h and cancel. Finally, r= 25-bh/ h