given $ \triangle a b c $ with vertices $ a(0, b), b(5, b + 1), c(6, b - 2) $ and $ \triangle d e f $ with vertices $ d(-b + 1,-4), e(-b, 1), f(-b + 3,2) $ prove $ \triangle a b c \cong \triangle d e f $ the lengths of the sides of $ \triangle a b c $ are $ a b= $ $ b c= $ and $ a c= $ the lengths of the sides of $ \triangle d e f $ are $ d e= $ $ e f= $ and $ d f= $ because $ a b=d e, b c=e f $, and $ a c=d f $, by the definition of congruent segments, $ \overline{a b} \cong \overline{d e}, \overline{b c} \cong \overline{e f} $, and $ \overline{a c} \cong \overline{d f} $. by the congruence theorem, $ \triangle a b c \cong \triangle d e f $.

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Accepted Answer

# Explanation:
## Step1: Calculate the length of $AB$
Use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. For $A(0,b)$ and $B(5,b + 1)$, we have $x_1 = 0,y_1 = b,x_2=5,y_2=b + 1$.
$$
\begin{align*}
AB&=\sqrt{(5 - 0)^2+[(b + 1)-b]^2}\
&=\sqrt{25+1}\
&=\sqrt{26}
\end{align*}
$$
## Step2: Calculate the length of $BC$
For $B(5,b + 1)$ and $C(6,b-2)$, $x_1 = 5,y_1 = b + 1,x_2=6,y_2=b-2$
$$
\begin{align*}
BC&=\sqrt{(6 - 5)^2+[(b-2)-(b + 1)]^2}\
&=\sqrt{1+(-3)^2}\
&=\sqrt{1 + 9}\
&=\sqrt{10}
\end{align*}
$$
## Step3: Calculate the length of $AC$
For $A(0,b)$ and $C(6,b-2)$, $x_1 = 0,y_1 = b,x_2=6,y_2=b-2$
$$
\begin{align*}
AC&=\sqrt{(6 - 0)^2+[(b-2)-b]^2}\
&=\sqrt{36+(-2)^2}\
&=\sqrt{36 + 4}\
&=\sqrt{40}=2\sqrt{10}
\end{align*}
$$
## Step4: Calculate the length of $DE$
For $D(-b + 1,-4)$ and $E(-b,1)$, $x_1=-b + 1,y_1=-4,x_2=-b,y_2=1$
$$
\begin{align*}
DE&=\sqrt{[-b-(-b + 1)]^2+(1-(-4))^2}\
&=\sqrt{(-1)^2+5^2}\
&=\sqrt{1+25}\
&=\sqrt{26}
\end{align*}
$$
## Step5: Calculate the length of $EF$
For $E(-b,1)$ and $F(-b + 3,2)$, $x_1=-b,y_1 = 1,x_2=-b + 3,y_2=2$
$$
\begin{align*}
EF&=\sqrt{[-b + 3-(-b)]^2+(2 - 1)^2}\
&=\sqrt{3^2+1^2}\
&=\sqrt{9+1}\
&=\sqrt{10}
\end{align*}
$$
## Step6: Calculate the length of $DF$
For $D(-b + 1,-4)$ and $F(-b + 3,2)$, $x_1=-b + 1,y_1=-4,x_2=-b + 3,y_2=2$
$$
\begin{align*}
DF&=\sqrt{[-b + 3-(-b + 1)]^2+(2-(-4))^2}\
&=\sqrt{2^2+6^2}\
&=\sqrt{4 + 36}\
&=\sqrt{40}=2\sqrt{10}
\end{align*}
$$

Since $AB = DE=\sqrt{26}$, $BC=EF=\sqrt{10}$, $AC = DF=2\sqrt{10}$, by the Side - Side - Side (SSS) Congruence Theorem, $\triangle ABC\cong\triangle DEF$

# Answer:
By calculating the lengths of the sides of $\triangle ABC$ and $\triangle DEF$ using the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$, we find that $AB = DE$, $BC=EF$, $AC = DF$. Then, by the Side - Side - Side (SSS) Congruence Theorem, $\triangle ABC\cong\triangle DEF$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate the length of \(AB\)

Step2: Calculate the length of \(BC\)

Step3: Calculate the length of \(AC\)

Step4: Calculate the length of \(DE\)

Step5: Calculate the length of \(EF\)

Step6: Calculate the length of \(DF\)

Answer: