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.
Proof with the help of Congruent Triangles
edit
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]
Proof with the help of Coordinate Geometry
edit
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]
