Theorem: Area of Triangle

This page gathers a variety of formulas for the area of a triangle.

Proof:
From Two-Point Form of Equation of Straight Line in Plane, the straight line $AB$ can be expressed as:
{{begin-eqn}} 
{{eqn | l = \dfrac {\paren {y_{AB} - y_1} } {\paren {x - x_1} }
      | r = \dfrac {\paren {y_2 - y_1} } {\paren {x_2 - x_1} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \dfrac {\paren {y_{AB} - v^2} } {\paren {x - v} }
      | r = \dfrac {\paren {u^2 - v^2} } {\paren {u - v} }
      | c = 
}}
{{eqn | r = \dfrac {\paren {u + v} \paren {u - v} } {\paren {u - v} }
      | c = Difference of Two Squares
}}
{{eqn | r = \paren {u + v}
      | c = dividing top and bottom by $\paren {u - v}$
}}
{{eqn | ll= \leadsto
      | l = \paren {y_{AB} - v^2}
      | r = \paren {u + v} \paren {x - v}
      | c = multiplying both sides by $\paren {x - v}$
}}
{{eqn | r = \paren {u + v} x - u v - v^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y_{AB}
      | r = \paren {u + v} x - u v
      | c = 
}}
{{end-eqn}} 
The coordinates of $C$ are: 
:$\tuple {\dfrac 1 2 \paren {u + v}, \dfrac 1 4 \paren {u + v}^2 } $
Let $P$ be the midpoint of $AB$, such that $AP = BP$.
We can use Two-Point Form of Equation of Straight Line in Plane above to determine the coordinates of $P$:
{{begin-eqn}} 
{{eqn | l = y_{AB}
      | r = \paren {u + v} x - u v
      | c = from above
}}
{{eqn | r = \paren {u + v} \paren {\dfrac 1 2 \paren {u + v} } - u v
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y
      | r = \dfrac 1 2 \paren {u^2 + v^2}
      | c = 
}}
{{end-eqn}} 
Therefore, the coordinates of $P$ are: 
:$\tuple {\dfrac 1 2 \paren {u + v}, \dfrac 1 2 \paren {u^2 + v^2} } $
:500px
We now connect points $P$ and $C$ to create line segment $PC$
This in turn creates two triangles.
We see from the diagram above:
:$\map \AA {\triangle ABC} = \map \AA {\triangle APC} + \map \AA {\triangle CPB}$
From Area of Triangle in Terms of Two Sides and Angle, we have:
:$\map \AA {\triangle APC} = \dfrac 1 2 \paren {AP} \paren {PC} \map \sin {\angle APC} $
We now notice that:
{{begin-eqn}} 
{{eqn | l = \paren {AP} \map \sin {\angle APC}
      | r = \dfrac {\paren {u + v} } 2 - v
      | c = from above
}}
{{eqn | r = \dfrac {\paren {u - v} } 2
      | c = 
}}
{{end-eqn}} 
Therefore:
:$\map \AA {\triangle APC} = \dfrac 1 2 \paren {PC}  \dfrac {\paren {u - v} } 2$
We also notice:
{{begin-eqn}} 
{{eqn | l = \map \sin {\angle CPB}
      | r = \map \sin {\pi - \angle APC}
      | c = $\angle CPB$ is supplementary to $\angle APC$
}}
{{eqn | r = \map \sin {\angle APC}
      | c = Sine of Supplementary Angle
}}
{{end-eqn}} 
Since $AP = PB$, $PC = PC$ and $\map \sin {\angle CPB} = \map \sin {\angle APC}$, then by Area of Triangle in Terms of Two Sides and Angle we have:
:$\map \AA {\triangle APC} = \map \AA {\triangle CPB}$
Therefore:
{{begin-eqn}} 
{{eqn | l = \map \AA {\triangle ABC} 
      | r = \map \AA {\triangle APC} + \map \AA {\triangle CPB}
      | c = 
}}
{{eqn | r = 2 \map \AA {\triangle APC} 
      | c = since $\map \AA {\triangle APC} = \map \AA {\triangle CPB}$
}}
{{eqn | r = 2 \dfrac 1 2 \paren {PC}  \dfrac {\paren {u - v} } 2
      | c = 
}}
{{eqn | r = \dfrac 1 2 \paren {u - v} P C
      | c = 
}}
{{end-eqn}} 
All that remains is to determine the length of $CP$:
{{begin-eqn}} 
{{eqn | l = CP
      | r = P_y - C_y
      | c = {{Defof|Length of Line}}
}}
{{eqn | l = CP
      | r = \dfrac 1 2 \paren {u^2 + v^2} - \dfrac 1 4 \paren {u + v}^2
      | c = substitution
}}
{{eqn | l = CP
      | r = \dfrac 1 4 \paren {2 u^2 + 2 v^2 - \paren {u + v}^2}
      | c = factor out $\dfrac 1 2$
}}
{{eqn | l = CP
      | r = \dfrac 1 4 \paren {u^2 - 2 u v + v^2}
      | c = expansion and subtraction
}}
{{eqn | l = CP
      | r = \dfrac 1 4 \paren {u - v}^2
      | c = factor
}}
{{end-eqn}} 
Substituting from above:
:$\map \AA {\triangle ABC} = \dfrac 1 2 \paren {u - v} \dfrac 1 4 \paren {u - v}^2$
Therefore:
:$\map \AA {\triangle ABC} = \dfrac 1 8 \paren {u - v}^3$
{{qed}}