c To prove our conjecture, it is enough to show that the corresponding sides in the triangles are congruent. We can do that using the Distance Formula.

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

Let's find the measure of

\overline{A B}

by substituting the coordinates of points

A

and

B

in the formula.

A B = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(- 3, - 5)$ & $(- 1, - 1)$

A B = (- 1 - (- 3))^{2} + (- 1 - (- 5))^{2}

▼

Simplify right-hand side

NegNeg

$- (- a) = a$

A B = (- 1 + 3)^{2} + (- 1 + 5)^{2}

AddTerms

Add terms

A B = 2^{2} + 4^{2}

CalcPow

Calculate power

A B = 4 + 16

AddTerms

Add terms

A B = 20

UseCalc

Use a calculator

A B = 4.472135 \dots

RoundDec

Round to $2$ decimal place(s)

A B \approx 4.47

Similarly, let's find the measure of

\overline{X Y} .

X Y = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

SubstitutePoints

Substitute $(5, - 5)$ & $(3, - 1)$

X Y = (3 - 5)^{2} + (- 1 - (- 5))^{2}

▼

Simplify right-hand side

NegNeg

$- (- a) = a$

X Y = (3 - 5)^{2} + (- 1 + 5)^{2}

AddSubTerms

Add and subtract terms

X Y = (- 2)^{2} + 4^{2}

CalcPow

Calculate power

X Y = 4 + 16

AddTerms

Add terms

X Y = 20

UseCalc

Use a calculator

X Y = 4.472135 \dots

RoundDec

Round to $2$ decimal place(s)

X Y \approx 4.47

Consequently, we have

\overline{A B} ≅ \overline{X Y} .

The lengths of the remaining sides are summarized in the table below.

Side	Distance	Measure
$\overline{A C}$	$A C = (- 1 - (- 3))^{2} + (- 5 - (- 5))^{2}$	$2$
$\overline{X Z}$	$X Z = (3 - 5)^{2} + (- 5 - (- 5))^{2}$	$2$
$\overline{B C}$	$B C = (- 1 - (- 1))^{2} + (- 5 - (- 1))^{2}$	$4$
$\overline{Y Z}$	$Y Z = (3 - 3)^{2} + (- 5 - (- 1))^{2}$	$4$

We got $\overline{A C} ≅ \overline{X Z}$ and $\overline{B C} ≅ \overline{Y Z} .$ Therefore, by the Side-Side-Side (SSS) Congruence Postulate we get $△ A B C ≅ △ X Y Z .$

Exercise 2 Page 268