Question

In ΔABC shown below, point A is at (0, 0), point B is at (x2, 0), point C is at (x1, y1), point D is at x sub 1 over 2, y sub 1 over 2, and point E is at the quantity of x sub 1 plus x sub 2 over 2, y sub 1 over 2:

How do I prove that segment DE is parallel to segment AB? 

Answer 1

you basically have to prove similarity, then use that to prove the two angles CDE & DAB are equal. You can prove similarity with AA, SSS, or SAS.
Hope it helps love.
-Joker7721

Answer 2

Answer with Step-by-step explanation:

We are given that a triangle ABC. In triangle ABC , the coordinates of A is at (0,0), Point B is at
`(x^2,0)`, Point C is at
`(x_1,y_1)`, Point D is at
`((x-1)/(2),(y-1)/(2))`, Point E is at
.

The coordinates of point E can be write as
`((3x-4)/(2).(y-1)/(2))`.

Slope formula of a line passing through two points
`(x_1,y_1)` and
`(x_2,y_2)`is given by

=
`(y_2-y_1)/(x_2-x_1)`.

By using slope formula of a line

Slope of line AB which passing through the points A (0,0) and B
`(x^2,0)`

Where
`x_1=0,x_2=x^2,y_1=0,y_2=0`

=
`(0-0)/(x^2-0)`

Slope of line AB which passing through the points A(0,0) and B
`(x^2,0),m_1`=0

Slope of line DE which passing through the points D
`((x-1)/(2),(y-1)/(2))` and E
`((3x-4)/(2),(y-1)/(2))`

Where
`x_1=(x-1)/(2),x_2=(3x-4)/(2),y_1=(y-1)/(2), y_2=(y-1)/(2)`

=
`((y-1)/(2)-(y-1)/(2))/((3x-4)/(2)-(x-1)/(2))`

Slope of line DE,
`m_2=((y-1-y+1)/(2))/((3x-4-x+1)/(2))`

Slope of line DE,
`m_2=(0)/((2x-3)/(2))`

Slope of line DE which passing through the points D
`((x-1)/(2),(y-1)/(2))`, and E
`((3x-4)/(2),(y-1)/(2))`

=0

We know that when two lines are parallel then thier slopes are equal.

`m_1=m_2`

Slope of line DE = Slope of line AB

Hence, the line DE and line AB are parallel lines.