1. in the diagram below, $\triangle def$ has been reflected in line $t$ by reflecting points $d$, $e$, and $f$
(a) as in class, draw in segments $\overline{dd}$, $\overline{ee}$ and $\overline{ff}$. mark the intersections of the segments with the line $t$.
(b) what is true about the points that you marked in part (a)? verify using an appropriate tool.
(c) what is true about the angles that the segments make with line $t$? mark these on your diagram.
(d) verify using tracing paper or a protractor that the two angles are equal.

2. obtuse triangle $\triangle rst$ is shown below along with line $m$. point $s$ lies on $m$. lines are drawn in from the vertices of $rst$ that are perpendicular to line $m$.
(a) using a compass only, draw the image of $\triangle rst$, i.e. $\triangle rst$ after a reflection in line $m$. leave all relevant construction marks.
(b) why are points $s$ and $s$ coincident (i.e. lie in the same spot)?
(c) draw segment $\overline{tt}$. why is $\triangle stt$ isosceles (if you dont remember what isosceles means, look it up).

Question

1. in the diagram below, $\triangle def$ has been reflected in line $t$ by reflecting points $d$, $e$, and $f$
   (a) as in class, draw in segments $\overline{dd}$, $\overline{ee}$ and $\overline{ff}$. mark the intersections of the segments with the line $t$.
   (b) what is true about the points that you marked in part (a)? verify using an appropriate tool.
   (c) what is true about the angles that the segments make with line $t$? mark these on your diagram.
   (d) verify using tracing paper or a protractor that the two angles are equal.

2. obtuse triangle $\triangle rst$ is shown below along with line $m$. point $s$ lies on $m$. lines are drawn in from the vertices of $rst$ that are perpendicular to line $m$.
   (a) using a compass only, draw the image of $\triangle rst$, i.e. $\triangle rst$ after a reflection in line $m$. leave all relevant construction marks.
   (b) why are points $s$ and $s$ coincident (i.e. lie in the same spot)?
   (c) draw segment $\overline{tt}$. why is $\triangle stt$ isosceles (if you dont remember what isosceles means, look it up).

Accepted Answer

### Problem 1 (a)
# Explanation:
## Step 1: Identify reflection points
For each point $ D $, $ E $, $ F $, their reflections are $ D' $, $ E' $, $ F' $ over line $ t $.
## Step 2: Draw segments
Draw $ \overline{DD'} $, $ \overline{EE'} $, $ \overline{FF'} $. The intersections of these segments with line $ t $ are the midpoints (let's call them $ M_D $, $ M_E $, $ M_F $ for $ \overline{DD'} $, $ \overline{EE'} $, $ \overline{FF'} $ respectively) since in a reflection, the line of reflection is the perpendicular bisector of the segment joining a point and its image.
(Note: Since this is a drawing step, the key is to connect each point to its image and mark where they cross line $ t $.)

### Problem 1 (b)
# Brief Explanations:
The marked points (intersections of $ \overline{DD'} $, $ \overline{EE'} $, $ \overline{FF'} $ with line $ t $) are the midpoints of each respective segment. To verify, we can use a ruler to measure the distance from each original point to the intersection point and from the intersection point to the image point—they should be equal. Also, the segments $ \overline{DD'} $, $ \overline{EE'} $, $ \overline{FF'} $ are perpendicular to line $ t $ (visible from the right angles in the diagram), so line $ t $ is the perpendicular bisector of each of these segments, meaning the intersection points are midpoints.
# Answer:
The marked points are the midpoints of $ \overline{DD'} $, $ \overline{EE'} $, and $ \overline{FF'} $ (line $ t $ is the perpendicular bisector of each segment).

### Problem 1 (c)
# Explanation:
## Step 1: Recall reflection properties
In a reflection over a line, the segments joining a point and its image (e.g., $ \overline{DD'} $, $ \overline{EE'} $, $ \overline{FF'} $) are perpendicular to the line of reflection $ t $, or make equal angles with line $ t $ (since they are perpendicular, the angles are $ 90^\circ $, or if we consider the angle between the segment and line $ t $, both sides of the line have equal angles).
## Step 2: Mark angles
Mark the right angles (or equal angles) between each segment ($ \overline{DD'} $, $ \overline{EE'} $, $ \overline{FF'} $) and line $ t $. From the diagram, we can see the right angle symbols, so the angles are $ 90^\circ $ (perpendicular) or the segments are perpendicular to line $ t $.
# Answer:
The segments $ \overline{DD'} $, $ \overline{EE'} $, $ \overline{FF'} $ are perpendicular to line $ t $ (the angles they make with line $ t $ are $ 90^\circ $, or equal on both sides of line $ t $).

### Problem 1 (d)
# Explanation:
## Step 1: Using tracing paper
Place tracing paper over the diagram, trace line $ t $ and one of the segments (e.g., $ \overline{DD'} $) and its angle with line $ t $. Then fold the tracing paper along line $ t $; the angle on one side should coincide with the angle on the other side, showing they are equal.
## Step 2: Using a protractor
Measure the angle between $ \overline{DD'} $ and line $ t $ on one side of line $ t $, then measure the angle on the other side. Both measurements should be equal (e.g., both $ 90^\circ $ if perpendicular, or equal acute/obtuse angles if not perpendicular, but in this case, they are perpendicular).
# Answer:
By tracing (folding along line $ t $ makes angles coincide) or protractor (measurements are equal), the two angles are equal.

### Problem 2 (a)
# Explanation:
## Step 1: Reflect point $ R $
Since we use a compass only, to reflect $ R $ over line $ m $:
- Draw a circle with center at the foot of the perpendicular from $ R $ to line $ m $ (let's call this foot $ P $) with radius $ RP $. The intersection of this circle with the other side of line $ m $ (along the perpendicular) is $ R' $.
- Point $ S $ is on line $ m $, so its reflection $ S' $ is $ S $ itself (since the distance from $ S $ to line $ m $ is $ 0 $).
- For point $ T $, draw a circle with center at the foot of the perpendicular from $ T $ to line $ m $ (foot $ Q $) with radius $ TQ $. The intersection on the other side of line $ m $ is $ T' $.
## Step 2: Connect $ R' $, $ S' $ (which is $ S $), and $ T' $
Draw $ \triangle R'S'T' $ by connecting these reflected points. Leave construction marks (arcs from the compass for the circles used to find $ R' $ and $ T' $).
(Note: This is a compass - only construction, so all steps use compass arcs to find the reflected points.)

### Problem 2 (b)
# Brief Explanations:
A reflection over a line maps a point to its image such that the line of reflection is the perpendicular bisector of the segment joining the point and its image. If a point lies on the line of reflection (line $ m $ in this case), the distance from the point to the line of reflection is $ 0 $. So, when we reflect point $ S $ over line $ m $, the image $ S' $ must also be at a distance of $ 0 $ from line $ m $, which means $ S' $ is the same as $ S $ (they coincide).
# Answer:
Point $ S $ lies on line $ m $ (the line of reflection). In a reflection, a point on the line of reflection maps to itself (since the distance from the point to the line is $ 0 $, so its image is also at distance $ 0 $ from the line, i.e., the same point). Thus, $ S $ and $ S' $ coincide.

### Problem 2 (c)
# Explanation:
## Step 1: Draw $ \overline{TT'} $
Using a straightedge, draw the segment connecting $ T $ and its reflection $ T' $ over line $ m $.
## Step 2: Analyze $ \triangle STT' $
Since $ T' $ is the reflection of $ T $ over line $ m $, line $ m $ is the perpendicular bisector of $ \overline{TT'} $. Let the foot of the perpendicular from $ T $ to line $ m $ be $ Q $. Then $ SQ $ is on line $ m $, and $ TQ = T'Q $ (because line $ m $ is the perpendicular bisector). Also, $ SQ $ is common to $ \triangle STQ $ and $ \triangle ST'Q $, and $ \angle SQT=\angle SQT' = 90^\circ $. By the Side - Angle - Side (SAS) congruence criterion, $ \triangle STQ\cong\triangle ST'Q $, so $ ST = ST' $. A triangle with two equal sides is isosceles, so $ \triangle STT' $ is isosceles.
# Answer:
After drawing $ \overline{TT'} $, $ \triangle STT' $ is isosceles because $ ST = ST' $ (since $ T' $ is the reflection of $ T $ over line $ m $, line $ m $ is the perpendicular bisector of $ \overline{TT'} $, so the distance from $ S $ to $ T $ equals the distance from $ S $ to $ T' $).

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QUESTION IMAGE

Question

Explanation:

Problem 1 (a)

Step 1: Identify reflection points

Step 2: Draw segments

Problem 1 (b)

Step 1: Recall reflection properties

Step 2: Mark angles

Answer:

Problem 1 (c)