Simplify the expression.
1 answer:
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CE ⊥ CF because m∠ECA + m∠FCA = 90°
Step-by-step explanation:
In ∆ABC
- On the extension of the side BC , draw a line segment CD ≅ CA
- Draw the segment AD
- The line segment CE is the angle bisector of ∠ACB
- The line segment CF is the median towards AD in ∆ ACD
We want to prove that CF ⊥ CE
Look to the attached figure
In Δ ABC
∵ CE is the bisector of angle ACB
∴ ∠ACE ≅ ∠BCE
In Δ ACD
∵ CA = CD
∴ Δ ACD is an isosceles triangle
∵ AD is the median towards AD
- In any isosceles triangle the median from a vertex to its opposite
side bisects this vertex
∴ AD bisects ∠ACD
∴ ∠ACF ≅ ∠DCF
∵ BCD is a straight segment
∵ CE , CA , CF are drawn from point C
∴ m∠BCE + m∠ACE + m∠ACF + m∠DCF = 180°
∵ m∠ACE ≅ m∠BCE
∵ m∠ACF ≅ m∠DCF
- Replace m∠BCE by m∠ACE and m∠DCF by m∠ACF
∴ m∠ACE + m∠ACE + m∠ACF + m∠ACF = 180°
∴ 2 m∠ACE + 2 m∠ACF = 180°
- Divide all terms by 2
∴ m∠ACE + m∠ACF = 90°
∴ EC ⊥ CF
CE ⊥ CF because m∠ECA + m∠FCA = 90°
Learn more:
You can learn more about perpendicular lines in brainly.com/question/11223427
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Let the point_1 = p₁ = (1,4)
and point_2 = p₂ = (-2,1)
and Point_3 = p₃ = (x,y)
The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)
The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d =
d₁ =
d₂ =
d₁ = d₂
∴
⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18 ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 =
∴ x = 4 or x = -2
∴ y = 1 or y = 7
Point_3 = (4,1) or (-2,7)
and point_2 = p₂ = (-2,1)
and Point_3 = p₃ = (x,y)
The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)
The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d =
d₁ =
d₂ =
d₁ = d₂
∴
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18 ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 =
∴ x = 4 or x = -2
∴ y = 1 or y = 7
Point_3 = (4,1) or (-2,7)
Answer:
Step-by-step explanation:
To solve this problem you must apply the proccedure shown below:
- By definition, the perimeter of the figure is the sum of the lenght of each side. Then, you must make the addition as following:
Where is the perimeter of the figure.
- Add the like terms, then, you obtain the following expression in simplest form: