reasoning
5. in the diagram shown, hk and ji intersect at point l such that jl ≅ kl and hl ≅ il.
(a) what rigid motion would map δhlk onto δilj? justify.
(b) why can we now conclude that hk must be congruent to ij?
6. in the diagram below, it is known that ∠cab ≅ ∠dab and ac ≅ ad (as marked).
(a) if point d was reflected across ab, explain why it must map onto point c.
(b) if δadb was reflected across ab, explain why it would map onto δacb. what does this mapping tell you about the two triangles?
(c) based on (b), why can we now claim that ab bisects ∠dbc?
(d) also based on (b), why can we now claim that point b is equidistant from points c and d?
n-gen math® geometry – unit 3 – euclidean triangle geometry – les (v10.15.24)
emathinstruction, red hook, ny 12571, © 2023

Question

Accepted Answer

### Problem 5(a)
# Explanation:
## Step1: Identify Rigid Motion
A 180 - degree rotation about point $ L $ will map $ \triangle HLK $ onto $ \triangle ILJ $. Justification: Given $ JL\cong KL $ and $ HL\cong IL $, and vertical angles $ \angle HLK $ and $ \angle ILJ $ are equal. A 180 - degree rotation about $ L $ will map $ H $ to $ I $, $ K $ to $ J $, and $ L $ to itself, so the triangles will coincide.
## Step2: Verify Congruence
After a 180 - degree rotation about $ L $, corresponding sides and angles of $ \triangle HLK $ and $ \triangle ILJ $ will be congruent, so the rotation is the rigid motion.
# Answer:
A 180 - degree rotation about point $ L $

### Problem 5(b)
# Explanation:
## Step1: Recall Rotation Property
Since $ \triangle HLK $ is mapped onto $ \triangle ILJ $ by a rigid motion (180 - degree rotation), rigid motions preserve congruence of corresponding sides.
## Step2: Identify Corresponding Sides
The side $ \overline{HK} $ in $ \triangle HLK $ corresponds to side $ \overline{IJ} $ in $ \triangle ILJ $ under the 180 - degree rotation about $ L $. So by the property of rigid motions (preservation of side lengths), $ \overline{HK}\cong\overline{IJ} $.
# Answer:
Because rigid motions (like the 180 - degree rotation) preserve side congruence, and $ \overline{HK} $ and $ \overline{IJ} $ are corresponding sides of congruent triangles (from the rotation), so $ \overline{HK}\cong\overline{IJ} $

### Problem 6(a)
# Explanation:
## Step1: Recall Reflection Properties
A reflection across a line (here $ \overline{AB} $) maps a point to another point such that the line is the perpendicular bisector of the segment joining the point and its image. Also, given $ \angle CAB\cong\angle DAB $ and $ \overline{AC}\cong\overline{AD} $, and $ \overline{AB} $ is common.
## Step2: Apply Reflection
When we reflect $ D $ across $ \overline{AB} $, the angle of incidence ( $ \angle DAB $) equals the angle of reflection ( $ \angle CAB $) and the distance from $ D $ to $ AB $ equals the distance from $ C $ to $ AB $. Also, since $ \overline{AC}\cong\overline{AD} $, the image of $ D $ across $ \overline{AB} $ must be $ C $ (because the reflection will map $ AD $ to $ AC $ as $ \angle DAB=\angle CAB $ and $ AD = AC $).
# Answer:
Because $ \angle CAB\cong\angle DAB $ (so the angle between $ AD $ and $ AB $ equals the angle between $ AC $ and $ AB $) and $ \overline{AC}\cong\overline{AD} $, reflecting $ D $ across $ \overline{AB} $ will map $ AD $ to $ AC $ and the perpendicular distance from $ D $ to $ AB $ equals that from $ C $ to $ AB $, so $ D $ maps to $ C $.

### Problem 6(b)
# Explanation:
## Step1: Reflect Triangle
Reflect $ \triangle ADB $ across $ \overline{AB} $. We know from part (a) that $ D $ maps to $ C $, $ A $ maps to $ A $ (since it's on the line of reflection), and $ B $ maps to $ B $ (also on the line of reflection).
## Step2: Determine Congruence
So the vertices of $ \triangle ADB $ ( $ A,D,B $) map to vertices of $ \triangle ACB $ ( $ A,C,B $) respectively. By the definition of congruent triangles (SSS or SAS, here $ AD = AC $, $ AB = AB $, and $ \angle DAB=\angle CAB $ so SAS), $ \triangle ADB\cong\triangle ACB $. This mapping tells us that $ \triangle ADB $ and $ \triangle ACB $ are congruent.
# Answer:
Reflecting $ \triangle ADB $ across $ \overline{AB} $ maps $ D $ to $ C $, $ A $ to $ A $, $ B $ to $ B $, so $ \triangle ADB $ maps onto $ \triangle ACB $. This tells us $ \triangle ADB\cong\triangle ACB $.

### Problem 6(c)
# Explanation:
## Step1: Use Congruent Triangles
From part (b), $ \triangle ADB\cong\triangle ACB $. Corresponding angles of congruent triangles are congruent, so $ \angle DBA\cong\angle CBA $.
## Step2: Define Angle Bisector
A ray (or line segment) that divides an angle into two congruent angles is an angle bisector. Since $ \angle DBA=\angle CBA $, $ \overline{AB} $ bisects $ \angle DBC $ (as $ \angle DBC=\angle DBA+\angle CBA $ and they are equal).
# Answer:
Since $ \triangle ADB\cong\triangle ACB $, $ \angle DBA=\angle CBA $, so $ \overline{AB} $ bisects $ \angle DBC $ (as it splits $ \angle DBC $ into two congruent angles).

### Problem 6(d)
# Explanation:
## Step1: Recall Congruent Triangles
From part (b), $ \triangle ADB\cong\triangle ACB $. Corresponding sides of congruent triangles are congruent, so $ \overline{BC}\cong\overline{BD} $.
## Step2: Define Equidistant
A point is equidistant from two other points if its distance to each point is equal. Since $ BC = BD $ (from $ \overline{BC}\cong\overline{BD} $), the distance from $ B $ to $ C $ equals the distance from $ B $ to $ D $, so $ B $ is equidistant from $ C $ and $ D $.
# Answer:
Since $ \triangle ADB\cong\triangle ACB $, $ \overline{BC}\cong\overline{BD} $, so $ B $ is equidistant from $ C $ and $ D $ (as $ BC = BD $).

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QUESTION IMAGE

Question

Explanation:

Problem 5(a)

Step1: Identify Rigid Motion

Step2: Verify Congruence

Step1: Recall Rotation Property

Step2: Identify Corresponding Sides

Step1: Recall Reflection Properties

Step2: Apply Reflection

Answer:

Problem 5(b)