In this right triangle,
A
D
=
C
D
,
B
E
=
C
E
,
D
E
=
1
2
A
B
{\displaystyle AD=CD,BE=CE,DE={\frac {1}{2}}AB}
D
E
∥
A
B
{\displaystyle DE\parallel AB}
The midpoint theorem is a fundamental concept in geometry that establishes a relationship between the midpoints of a triangle's sides. This theorem states that when you connect the midpoints of two sides of a triangle, the resulting line segment is parallel to the third side. Additionally, this line segment is precisely half the length of the third side.
In a triangle, if a line segment connects the midpoints of two sides, then this line segment is parallel to the third side and half its length.
The construction for the mid-point theorem's proof with similar triangles Proposition: Let
D
{\displaystyle D}
E
{\displaystyle E}
A
C
{\displaystyle AC}
B
C
{\displaystyle BC}
A
B
C
{\displaystyle ABC}
D
E
∥
A
B
{\displaystyle DE\parallel AB}
D
E
=
1
2
A
B
{\displaystyle DE={\frac {1}{2}}AB}
Construction: Add
D
{\displaystyle D}
E
{\displaystyle E}
D
E
{\displaystyle DE}
F
{\displaystyle F}
E
F
=
D
E
{\displaystyle EF=DE}
B
{\displaystyle B}
F
{\displaystyle F}
Proof: [1]
Δ
C
D
E
{\displaystyle \Delta CDE}
Δ
E
B
F
,
{\displaystyle \Delta EBF,}
C
E
=
B
E
{\displaystyle CE=BE}
D
E
=
E
F
{\displaystyle DE=EF}
∠
C
E
D
=
∠
A
E
F
{\displaystyle \angle CED=\angle AEF}
∴
Δ
C
D
E
≅
Δ
E
B
F
{\displaystyle \Delta CDE\cong \Delta EBF}
So,
∠
C
D
E
=
∠
B
F
E
{\displaystyle \angle CDE=\angle BFE}
∴
C
D
∥
B
F
{\displaystyle CD\parallel BF}
Or,
A
D
∥
B
F
{\displaystyle AD\parallel BF}
C
D
=
B
F
=
D
A
{\displaystyle CD=BF=DA}
Therefore,
A
D
F
B
{\displaystyle ADFB}
∴
D
F
∥
A
B
{\displaystyle DF\parallel AB}
D
E
∥
A
B
{\displaystyle DE\parallel AB}
[2]
D
F
=
A
B
{\displaystyle DF=AB}
Or
D
E
+
E
F
=
A
B
{\displaystyle DE+EF=AB}
Or,
D
E
+
D
E
=
A
B
{\displaystyle DE+DE=AB}
Δ
C
D
E
≅
Δ
E
B
F
{\displaystyle \Delta CDE\cong \Delta EBF}
Or,
2
D
E
=
A
B
{\displaystyle 2DE=AB}
Or,
D
E
=
1
2
A
B
{\displaystyle DE={\frac {1}{2}}AB}
∴ In the triangle
Δ
A
B
C
,
{\displaystyle \Delta ABC,}
D
E
∥
A
B
{\displaystyle DE\parallel AB}
D
E
=
1
2
A
B
{\displaystyle DE={\frac {1}{2}}AB}
D
{\displaystyle D}
E
{\displaystyle E}
A
C
{\displaystyle AC}
B
C
{\displaystyle BC}
[Proved]
Proposition: Let
D
{\displaystyle D}
E
{\displaystyle E}
A
C
{\displaystyle AC}
A
B
{\displaystyle AB}
A
B
C
{\displaystyle ABC}
A
,
B
,
C
{\displaystyle A,B,C}
A
(
x
1
,
y
1
)
,
B
(
x
2
,
y
2
)
,
C
(
x
3
,
y
3
)
{\displaystyle A(x_{1},y_{1}),B(x_{2},y_{2}),C(x_{3},y_{3})}
D
E
=
1
2
B
C
{\displaystyle DE={\frac {1}{2}}BC}
D
E
∥
B
C
{\displaystyle DE\parallel BC}
Proof: [1]
B
C
=
(
x
3
−
x
2
)
2
+
(
y
3
−
y
2
)
2
{\displaystyle BC={\sqrt {(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}}}
The midpoint of
A
(
x
1
,
y
1
)
{\displaystyle A(x_{1},y_{1})}
C
(
x
3
,
y
3
)
{\displaystyle C(x_{3},y_{3})}
D
(
x
1
+
x
3
2
,
y
1
+
y
3
2
)
{\displaystyle D({\frac {x_{1}+x_{3}}{2}},{\frac {y_{1}+y_{3}}{2}})}
In the same way, The midpoint of
A
(
x
1
,
y
1
)
{\displaystyle A(x_{1},y_{1})}
B
(
x
2
,
y
2
)
{\displaystyle B(x_{2},y_{2})}
E
(
x
1
+
x
2
2
,
y
1
+
y
2
2
)
{\displaystyle E({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}})}
∴ The distance of
D
E
=
(
x
1
+
x
3
2
−
x
1
+
x
2
2
)
2
+
(
y
1
+
y
3
2
−
y
1
+
y
2
2
)
2
{\displaystyle DE={\sqrt {({\frac {x_{1}+x_{3}}{2}}-{\frac {x_{1}+x_{2}}{2}})^{2}+({\frac {y_{1}+y_{3}}{2}}-{\frac {y_{1}+y_{2}}{2}})^{2}}}}
=
(
x
1
+
x
3
−
x
1
−
x
2
2
)
2
+
(
y
1
+
y
3
−
y
1
−
y
2
2
)
2
{\displaystyle ={\sqrt {({\frac {x_{1}+x_{3}-x_{1}-x_{2}}{2}})^{2}+({\frac {y_{1}+y_{3}-y_{1}-y_{2}}{2}})^{2}}}}
=
(
x
3
−
x
2
)
2
4
+
(
y
3
−
y
2
)
2
4
{\displaystyle ={\sqrt {{\frac {(x_{3}-x_{2})^{2}}{4}}+{\frac {(y_{3}-y_{2})^{2}}{4}}}}}
=
(
x
3
−
x
2
)
2
+
(
y
3
−
y
2
)
2
4
{\displaystyle ={\sqrt {\frac {(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}{4}}}}
=
(
x
3
−
x
2
)
2
+
(
y
3
−
y
2
)
2
2
{\displaystyle ={\frac {\sqrt {(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}}{2}}}
=
1
2
B
C
{\displaystyle ={\frac {1}{2}}BC}
B
C
=
(
x
3
−
x
2
)
2
+
(
y
3
−
y
2
)
2
{\displaystyle BC={\sqrt {(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}}}
[2]
B
C
,
{\displaystyle BC,}
m
1
=
y
2
−
y
3
x
2
−
x
3
{\displaystyle m_{1}={\frac {y_{2}-y_{3}}{x_{2}-x_{3}}}}
The slope of
D
E
,
{\displaystyle DE,}
m
2
=
y
1
+
y
2
2
−
y
1
+
y
3
2
x
1
+
x
2
2
−
x
1
+
x
3
2
{\displaystyle m_{2}={\frac {{\frac {y_{1}+y_{2}}{2}}-{\frac {y_{1}+y_{3}}{2}}}{{\frac {x_{1}+x_{2}}{2}}-{\frac {x_{1}+x_{3}}{2}}}}}
=
y
1
+
y
2
−
y
1
−
y
3
2
x
1
+
x
2
−
x
1
−
x
3
2
{\displaystyle ={\frac {\frac {y_{1}+y_{2}-y_{1}-y_{3}}{2}}{\frac {x_{1}+x_{2}-x_{1}-x_{3}}{2}}}}
=
y
2
−
y
3
2
x
2
−
x
3
2
{\displaystyle ={\frac {\frac {y_{2}-y_{3}}{2}}{\frac {x_{2}-x_{3}}{2}}}}
=
y
2
−
y
3
x
2
−
x
3
{\displaystyle ={\frac {y_{2}-y_{3}}{x_{2}-x_{3}}}}
=
m
1
{\displaystyle =m_{1}}
m
1
=
y
2
−
y
3
x
2
−
x
3
{\displaystyle m_{1}={\frac {y_{2}-y_{3}}{x_{2}-x_{3}}}}
Therefore,
D
E
∥
B
C
{\displaystyle DE\parallel BC}
∴ In the triangle
Δ
A
B
C
,
{\displaystyle \Delta ABC,}
D
E
∥
B
C
{\displaystyle DE\parallel BC}
D
E
=
1
2
B
C
{\displaystyle DE={\frac {1}{2}}BC}
D
{\displaystyle D}
E
{\displaystyle E}
A
C
{\displaystyle AC}
A
B
{\displaystyle AB}
[Proved]
