# Which property, postulate, or definition justifies the statement below? If KL = MN and KL + PQ = RS , then MN + PQ = RS

###### Question:

## Answers

huh im so confused

Step-by-step explanation:

Answer with explanation:

The two Starting Points used by Eowyn for deriving Distance formula:

[tex]=A(x_{1},y_{1}) \text{and} B(x_{2},y_{2})[/tex]

Then Eowyn has Drawn the point C

→ She Draw a Horizontal Line From Point A, and a vertical line from point B, and then the point where these lines intersect was named the point [tex]C(x_{3},y_{3})[/tex].

→ She Supposed, AC=b, BC=a

and then, AB=c

→Ruler Postulate

[tex]a=|x_{2}-x_{1}|\\\\b=|y_{2}-y_{1}|[/tex]

With the Help of Pythagorean Theorem and Substitution Property

[tex]c^2=a^2+b^2\\\\AB^2=AC^2+BC^2\\\\AB^2=(x_{2}-x_{1})^2+(y_{2}-y_{1})^2\\\\AB=\pm \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]

As, Distance can't be negative

So, [tex]AB= \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]

⇒ Here, a and b should be

[tex]a=|y_{2}-y_{1}|,b=|x_{2}-x_{1}|[/tex]

Option C:→→ Eowyn didn't correctly derive the distance formula. She drew in her auxiliary segments correctly, but she didn't define a, b, and c correctly in order to apply them in the Pythagorean Theorem.

[tex]When deriving the distance formula, eowyn starts with two points, a(x1,y1) and b(x2,y2). she then dr[/tex]

E

Step-by-step explanation:

Shapes that have perpendicular diagonals (including squares) have perpendicular diagonals.

Step-by-step explanation:

the answer is a because it is thanks i hope you get it right

Additive property of equality states that:

if a=b, then a+c = b+c,

that is, if 2 quantities of the same value are added the same quantity, they are again equal.

In our problem a=AD, b=DB, and c= CD=DE

thus a+c=b+c means AD + CD = DE + DE,

by the Addition Property

Addition Property