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 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 , Point C is at , Point D is at , Point E is at .

The coordinates of point E can be write as .

Slope formula of a line passing through two points and is given by

=.

By using slope formula of a line

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

Where

=

Slope of line AB which passing through the points A(0,0) and B=0

Slope of line DE which passing through the points D and E

Where

=

Slope of line DE,

Slope of line DE,

Slope of line DE which passing through the points D, and E

=0

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

Slope of line DE = Slope of line AB

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