Question

1. derek is proving that any point on the perpendicular bisector of a line segment is equidistant from the endpoints of the line segment. he is using the figure below. he knows that line ab is the perpendicular bisector of cd. finish the proof he started below.

statement\treason
ab is perpendicular bisector of cd\tgiven
draw ab and ad\tthere exists a line through any 2 points

1. two triangles, △abc and △def, are placed on a coordinate plane. the coordinates are:

● a(0, 0), b(4, 0), c(2, 3)
● d(1, 1), e(5, 1), f(3, 4)
a. a student claims the triangles are congruent because they “look the same.” use a rigid motion (translation, reflection, or rotation) to justify whether △abc≅△def. explain your reasoning.
b. another student claims △abc≅△def by sss congruence. evaluate whether this reasoning is valid. show calculations to support your conclusion.

Explanation:

7.

Step1: Define perpendicular and bisector properties

Since $\overline{AB}$ is the perpendicular bisector of $\overline{CD}$, $\angle ABD=\angle ABC = 90^{\circ}$ and $BD = BC$ (Definition of perpendicular bisector).

Step2: Consider common side

$\overline{AB}=\overline{AB}$ (Reflexive property of congruence).

Step3: Prove triangle congruence

$\triangle ABD\cong\triangle ABC$ by the Side - Angle - Side (SAS) congruence criterion.

Step4: Use congruent - triangle properties

Since $\triangle ABD\cong\triangle ABC$, $AD = AC$ (Corresponding parts of congruent triangles are congruent). So any point $A$ on the perpendicular bisector $\overline{AB}$ of $\overline{CD}$ is equidistant from the endpoints $C$ and $D$.

Statement	Reason
$\overline{AB}=\overline{AB}$	Reflexive property
$\triangle ABD\cong\triangle ABC$	SAS congruence criterion
$AD = AC$	Corresponding parts of congruent triangles are congruent

8. A.

Step1: Find the translation rule

To check if we can use a rigid - motion, consider a translation. The $x$ - coordinate of $A(0,0)$ and $D(1,1)$. To get from $A$ to $D$, we have the translation rule $(x,y)\to(x + 1,y + 1)$.
Apply this rule to $B(4,0)$: $(4,0)\to(4 + 1,0+1)=(5,1)$ which is the coordinates of $E$.
Apply this rule to $C(2,3)$: $(2,3)\to(2 + 1,3 + 1)=(3,4)$ which is the coordinates of $F$.
Since we can translate $\triangle ABC$ to $\triangle DEF$ using the rule $(x,y)\to(x + 1,y + 1)$, the two triangles are congruent by the definition of congruence (a rigid motion that maps one triangle onto the other).

Step1: Use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$

For side lengths of $\triangle ABC$:
$AB=\sqrt{(4 - 0)^2+(0 - 0)^2}=\sqrt{4^2}=4$
$BC=\sqrt{(2 - 4)^2+(3 - 0)^2}=\sqrt{(-2)^2+3^2}=\sqrt{4 + 9}=\sqrt{13}$
$AC=\sqrt{(2 - 0)^2+(3 - 0)^2}=\sqrt{2^2+3^2}=\sqrt{4 + 9}=\sqrt{13}$

For side lengths of $\triangle DEF$:
$DE=\sqrt{(5 - 1)^2+(1 - 1)^2}=\sqrt{4^2}=4$
$EF=\sqrt{(3 - 5)^2+(4 - 1)^2}=\sqrt{(-2)^2+3^2}=\sqrt{4 + 9}=\sqrt{13}$
$DF=\sqrt{(3 - 1)^2+(4 - 1)^2}=\sqrt{2^2+3^2}=\sqrt{4 + 9}=\sqrt{13}$

Since $AB = DE = 4$, $BC=EF=\sqrt{13}$, and $AC = DF=\sqrt{13}$, by the SSS (Side - Side - Side) congruence criterion, $\triangle ABC\cong\triangle DEF$.

Answer:

The triangles are congruent because $\triangle ABC$ can be translated to $\triangle DEF$ using the rule $(x,y)\to(x + 1,y + 1)$.

8. B.