# Lesson 10

Practicing Proofs

### Problem 1

Painters and carpenters use scaffolding to climb buildings from the outside. What shapes do you see? Why does one figure have more right angles?

### Problem 2

Select all true statements based on the diagram.

A:

Angle $$CBE$$ is congruent to angle $$ABE$$.

B:

Angle $$CEB$$ is congruent to angle $$DEA$$.

C:

Segment $$DA$$ is congruent to segment $$CB$$.

D:

Segment $$DC$$ is congruent to segment $$AB$$.

E:

Line $$DC$$ is parallel to line $$AB$$.

F:

Line $$DA$$ is parallel to line $$CB$$.

### Problem 3

Prove $$ABCD$$ is a parallelogram.

### Problem 4

Tyler has proven that triangle $$WYZ$$ is congruent to triangle $$WYX$$ using the Side-Side-Side Triangle Congruence Theorem. Why can he now conclude that diagonal $$WY$$ bisects angles $$ZWX$$ and $$ZYX$$?

### Solution

(From Unit 2, Lesson 9.)

### Problem 5

$$WXYZ$$ is a kite. Angle $$WXY$$ has a measure of 133 degrees and angle $$ZYX$$ has a measure of 34 degrees. Find the measure of angle $$ZWY$$.

### Solution

(From Unit 2, Lesson 9.)

### Problem 6

Elena is thinking through a proof using a reflection to show that the base angles of an isosceles triangle are congruent. Complete the missing information for her proof.

Call the midpoint of segment $$CD$$ $$\underline{\hspace{.5in}1\hspace{.5in}}$$. Construct the perpendicular bisector of segment $$CD$$. The perpendicular bisector of $$CD$$ must go through $$B$$ since it's the midpoint. $$A$$ is also on the perpendicular of $$CD$$ because the distance from $$A$$ to $$\underline{\hspace{.5in}2\hspace{.5in}}$$ is the same as the distance from $$A$$ to $$\underline{\hspace{.5in}3\hspace{.5in}}$$. We want to show triangle $$ADC$$ is congruent to triangle $$ACD$$. Reflect triangle $$ADC$$ across line $$\underline{\hspace{.5in}4\hspace{.5in}}$$. Since $$\underline{\hspace{.5in}5\hspace{.5in}}$$ is on the line of reflection, it definitely lines up with itself. $$DB$$ is congruent to $$\underline{\hspace{.5in}6\hspace{.5in}}$$ since $$AB$$ is the perpendicular bisector of $$CD$$. $$D’$$ will coincide with $$\underline{\hspace{.5in}7\hspace{.5in}}$$ since it is on the other side of a perpendicular line and the same distance from it (and that’s the definition of reflection!). $$C’$$ will coincide with $$\underline{\hspace{.5in}8\hspace{.5in}}$$ since it is on the other side of a perpendicular line and the same distance from it (and that’s the definition of reflection!). Since the rigid transformation will take triangle $$ADC$$ onto triangle $$ACD$$, that means angle $$\underline{\hspace{.5in}9\hspace{.5in}}$$ will be taken onto angle $$\underline{\hspace{.5in}10\hspace{.5in}}$$ (they are corresponding parts under the same reflection), and therefore they are congruent.

### Solution

(From Unit 2, Lesson 8.)

### Problem 7

Segment $$EG$$ is an angle bisector of angle $$FGH$$. Noah wrote a proof to show that triangle $$HEG$$ is congruent to triangle $$FEG$$. Noah's proof is not correct. Why is Noah's proof incorrect?

​​​​​​

• Side $$EG$$ is congruent to side $$EG$$ because they're the same segment.
• Angle $$EGH$$ is congruent to angle $$EGF$$ because segment $$EG$$ is an angle bisector of angle $$FGH$$.
• Angle $$HEG$$ is congruent to angle $$FEG$$ because segment $$EG$$ is an angle bisector of angle $$FGH$$.
• By the Angle-Side-Angle Triangle Congruence Theorem, triangle $$HEG$$ is congruent to triangle $$FEG$$.

### Solution

(From Unit 2, Lesson 7.)

### Problem 8

Figure $$HNMLKEFG$$ is the image of figure $$ABCDKLMN$$ after being rotated 90 degrees counterclockwise around point $$K$$. Draw an auxiliary line in figure $$ABCDKLMN$$ to create a quadrilateral. Draw the image of the auxiliary line when rotated 90 degrees counterclockwise around point $$K$$

Write a congruence statement for the quadrilateral you created in figure $$ABCDKLMN$$ and the image of the quadrilateral in figure $$HNMLKEFG$$.