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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? 




In ΔABC shown below, point A is at (0, 0), point B is at (x2, 0), point C is at (x-example-1
User Borja
by
7.9k points

2 Answers

2 votes
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
User Patroclus
by
8.5k points
4 votes

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
(\frac{x-1(x-2)/(2),(y-1)/(2)).

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.

User James Alvarez
by
8.1k points
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