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. Completing the proof

Step1: Define mid - point property

Since $\overline{AB}$ is the perpendicular bisector of $\overline{CD}$, let the intersection of $\overline{AB}$ and $\overline{CD}$ be point $O$. Then $CO = OD$ (Definition of a bisector).
Statement: $CO=OD$; Reason: Definition of a bisector

Step2: Right - angle property

$\angle AOC=\angle AOD = 90^{\circ}$ (Definition of perpendicular lines).
Statement: $\angle AOC=\angle AOD = 90^{\circ}$; Reason: Definition of perpendicular lines

Step3: Common side

$AO = AO$ (Reflexive property of equality).
Statement: $AO = AO$; Reason: Reflexive property of equality

Step4: Prove triangle congruence

$\triangle AOC\cong\triangle AOD$ by the Side - Angle - Side (SAS) congruence criterion.
Statement: $\triangle AOC\cong\triangle AOD$; Reason: SAS congruence criterion

Step5: Corresponding parts of congruent triangles

$AC = AD$ (Corresponding parts of congruent triangles are equal).
Statement: $AC = AD$; Reason: Corresponding parts of congruent triangles are equal

8.

A.

Translate $\triangle ABC$ one unit to the right and one unit up.
The translation rule is $(x,y)\to(x + 1,y + 1)$.
For point $A(0,0)$, $A'(0+1,0 + 1)=(1,1)$; for point $B(4,0)$, $B'(4 + 1,0+1)=(5,1)$; for point $C(2,3)$, $C'(2+1,3 + 1)=(3,4)$.
After translation, the vertices of $\triangle ABC$ match the vertices of $\triangle DEF$. So, $\triangle ABC\cong\triangle DEF$ by translation.

Use the distance formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$ to find the side - lengths of the triangles.
For $\triangle ABC$:
$AB=\sqrt{(4 - 0)^2+(0 - 0)^2}=\sqrt{16}=4$
$BC=\sqrt{(2 - 4)^2+(3 - 0)^2}=\sqrt{(-2)^2+3^2}=\sqrt{4 + 9}=\sqrt{13}$
$AC=\sqrt{(2 - 0)^2+(3 - 0)^2}=\sqrt{4 + 9}=\sqrt{13}$

For $\triangle DEF$:
$DE=\sqrt{(5 - 1)^2+(1 - 1)^2}=\sqrt{16}=4$
$EF=\sqrt{(3 - 5)^2+(4 - 1)^2}=\sqrt{(-2)^2+3^2}=\sqrt{4 + 9}=\sqrt{13}$
$DF=\sqrt{(3 - 1)^2+(4 - 1)^2}=\sqrt{4 + 9}=\sqrt{13}$

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

Answer:

$\triangle ABC$ can be translated one unit to the right and one unit up to coincide with $\triangle DEF$. So, $\triangle ABC\cong\triangle DEF$.

B.