Answer:
The line containing the midpoints of two sides of a triangle is parallel to the third side ⇒ proved down
Step-by-step explanation:
* Lets revise the rules of the midpoint and the slope to prove the
problem
- The slope of a line whose endpoints are (x1 , y1) and (x2 , y2) is
[tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
- The mid-point of a line whose endpoints are (x1 , y1) and (x2 , y2) is
[tex](\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})[/tex]
* Lets solve the problem
- PQR is a triangle of vertices P (0 , 0) , Q (2a , 0) , R (2b , 2c)
- Lets find the mid-poits of PQ called A
∵ Point P is (x1 , y1) and point Q is (x2 , y2)
∴ x1 = 0 , x2 = 2a and y1 = 0 , y2 = 0
∵ A is the mid-point of PQ
∴ [tex]A=(\frac{0+2a}{2},\frac{0+0}{2})=(\frac{2a}{2},\frac{0}{2})=(a,0)[/tex]
- Lets find the mid-poits of PR which called B
∵ Point P is (x1 , y1) and point R is (x2 , y2)
∴ x1 = 0 , x2 = 2b and y1 = 0 , y2 = 2c
∵ B is the mid-point of PR
∴ [tex]B=(\frac{0+2b}{2},\frac{0+2c}{2})=(\frac{2b}{2},\frac{2c}{2})=(b,c)[/tex]
- The parallel line have equal slopes, so lets find the slopes of AB and
QR to prove that they have same slopes then they are parallel
# Slope of AB
∵ Point A is (x1 , y1) and point B is (x2 , y2)
∵ Point A = (a , 0) and point B = (b , c)
∴ x1 = a , x2 = b and y1 = 0 and y2 = c
∴ The slope of AB is [tex]m=\frac{c-0}{b-a}=\frac{c}{b-a}[/tex]
# Slope of QR
∵ Point Q is (x1 , y1) and point R is (x2 , y2)
∵ Point Q = (2a , 0) and point R = (2b , 2c)
∴ x1 = 2a , x2 = 2b and y1 = 0 and y2 = 2c
∴ The slope of AB is [tex]m=\frac{2c-0}{2b-2a}=\frac{2c}{2(b-c)}=\frac{c}{b-a}[/tex]
∵ The slopes of AB and QR are equal
∴ AB // QR
∵ AB is the line containing the midpoints of PQ and PR of Δ PQR
∵ QR is the third side of the triangle
∴ The line containing the midpoints of two sides of a triangle is parallel
to the third side
