Question
The swan below is composed of several triangles. Use the given information and the figure to find each angle measure. Note: Figure not drawn to scale.
Given: ΔABC is equilateral; DE¯¯¯¯¯¯¯¯≅CE¯¯¯¯¯¯¯¯≅EF¯¯¯¯¯¯¯¯
\overline{DE}\cong\overline{CE}\cong\overline{EF}; KO¯¯¯¯¯¯¯¯≅FO¯¯¯¯¯¯¯¯
\overline{KO}\cong\overline{FO}; JN¯¯¯¯¯¯¯¯≅JO¯¯¯¯¯¯¯
\overline{JN}\cong\overline{JO}; ∠BCD ≅ ∠BDC; CGF ≅ ∠CFG; ∠HKN ≅ ∠HNK; ΔGCF ≅ ΔGKF ≅ ΔJHM; ΔKFH ≅ ΔKLH
Triangular Swan Portfolio
Print the document so that you can write on it and mark the pictureshowin congruent sides
(2 points)
1. m∠ABC =
2. m∠BCA =
3. m∠CAB =
4. m∠BCD =
5. m∠BDC =
6. m∠CBD =
Question 3
(2 points)
7. m∠EDC =
8. m∠ECD =
9. m∠CED =
10. m∠ECF =
11. m∠EFC =
12. m∠CEF =
Given: ΔABC is equilateral; DE¯¯¯¯¯¯¯¯≅CE¯¯¯¯¯¯¯¯≅EF¯¯¯¯¯¯¯¯
\overline{DE}\cong\overline{CE}\cong\overline{EF}; KO¯¯¯¯¯¯¯¯≅FO¯¯¯¯¯¯¯¯
\overline{KO}\cong\overline{FO}; JN¯¯¯¯¯¯¯¯≅JO¯¯¯¯¯¯¯
\overline{JN}\cong\overline{JO}; ∠BCD ≅ ∠BDC; CGF ≅ ∠CFG; ∠HKN ≅ ∠HNK; ΔGCF ≅ ΔGKF ≅ ΔJHM; ΔKFH ≅ ΔKLH
Triangular Swan Portfolio
Print the document so that you can write on it and mark the pictureshowin congruent sides
(2 points)
1. m∠ABC =
2. m∠BCA =
3. m∠CAB =
4. m∠BCD =
5. m∠BDC =
6. m∠CBD =
Question 3
(2 points)
7. m∠EDC =
8. m∠ECD =
9. m∠CED =
10. m∠ECF =
11. m∠EFC =
12. m∠CEF =
Answers
GPT-4o mini
To solve the problem, let's analyze the given information step by step.
### Step 1: Analyze Triangle ABC
Since \( \Delta ABC \) is equilateral:
- All angles are equal.
The measure of each angle in an equilateral triangle is given by:
\[
m\angle ABC = m\angle BCA = m\angle CAB = \frac{180^\circ}{3} = 60^\circ
\]
**So:**
1. \( m\angle ABC = 60^\circ \)
2. \( m\angle BCA = 60^\circ \)
3. \( m\angle CAB = 60^\circ \)
### Step 2: Analyze Angles BCD and BDC
Given that \( \angle BCD \cong \angle BDC \), we can denote:
\[
\angle BCD = \angle BDC = x
\]
From the triangle sum theorem:
\[
m\angle ABC + m\angle BCD + m\angle BDC = 180^\circ
\]
Substituting the known values:
\[
60^\circ + x + x = 180^\circ
\]
\[
60^\circ + 2x = 180^\circ
\]
\[
2x = 120^\circ
\]
\[
x = 60^\circ
\]
Thus:
4. \( m\angle BCD = 60^\circ \)
5. \( m\angle BDC = 60^\circ \)
6. Since \( m\angle CBD = 180^\circ - m\angle ABC - m\angle BDC \):
\[
m\angle CBD = 180^\circ - 60^\circ - 60^\circ = 60^\circ
\]
### Step 3: Analyze Triangle CDE and Triangle ECF
Regarding triangle \( CDE \):
Given that \( DE \cong CE \cong EF \), triangle \( CDE \) is also equilateral. Therefore:
7. \( m\angle EDC = 60^\circ \)
8. \( m\angle ECD = 60^\circ \)
9. \( m\angle CED = 60^\circ \)
For triangle \( ECF \) (since \( ECF \) is not specified as equilateral, but given the congruency relationships and symmetry):
10. \( m\angle ECF = 60^\circ \)
11. \( m\angle EFC = 60^\circ \)
12. Similarly, \( m\angle CEF = 60^\circ \)
### Summary of Angles
1. \( m\angle ABC = 60^\circ \)
2. \( m\angle BCA = 60^\circ \)
3. \( m\angle CAB = 60^\circ \)
4. \( m\angle BCD = 60^\circ \)
5. \( m\angle BDC = 60^\circ \)
6. \( m\angle CBD = 60^\circ \)
7. \( m\angle EDC = 60^\circ \)
8. \( m\angle ECD = 60^\circ \)
9. \( m\angle CED = 60^\circ \)
10. \( m\angle ECF = 60^\circ \)
11. \( m\angle EFC = 60^\circ \)
12. \( m\angle CEF = 60^\circ \)
All of the angles found in triangles \( ABC \), \( CDE \), and \( ECF \) are \( 60^\circ \) due to the properties of equilateral triangles and the given conditions of congruence.
### Step 1: Analyze Triangle ABC
Since \( \Delta ABC \) is equilateral:
- All angles are equal.
The measure of each angle in an equilateral triangle is given by:
\[
m\angle ABC = m\angle BCA = m\angle CAB = \frac{180^\circ}{3} = 60^\circ
\]
**So:**
1. \( m\angle ABC = 60^\circ \)
2. \( m\angle BCA = 60^\circ \)
3. \( m\angle CAB = 60^\circ \)
### Step 2: Analyze Angles BCD and BDC
Given that \( \angle BCD \cong \angle BDC \), we can denote:
\[
\angle BCD = \angle BDC = x
\]
From the triangle sum theorem:
\[
m\angle ABC + m\angle BCD + m\angle BDC = 180^\circ
\]
Substituting the known values:
\[
60^\circ + x + x = 180^\circ
\]
\[
60^\circ + 2x = 180^\circ
\]
\[
2x = 120^\circ
\]
\[
x = 60^\circ
\]
Thus:
4. \( m\angle BCD = 60^\circ \)
5. \( m\angle BDC = 60^\circ \)
6. Since \( m\angle CBD = 180^\circ - m\angle ABC - m\angle BDC \):
\[
m\angle CBD = 180^\circ - 60^\circ - 60^\circ = 60^\circ
\]
### Step 3: Analyze Triangle CDE and Triangle ECF
Regarding triangle \( CDE \):
Given that \( DE \cong CE \cong EF \), triangle \( CDE \) is also equilateral. Therefore:
7. \( m\angle EDC = 60^\circ \)
8. \( m\angle ECD = 60^\circ \)
9. \( m\angle CED = 60^\circ \)
For triangle \( ECF \) (since \( ECF \) is not specified as equilateral, but given the congruency relationships and symmetry):
10. \( m\angle ECF = 60^\circ \)
11. \( m\angle EFC = 60^\circ \)
12. Similarly, \( m\angle CEF = 60^\circ \)
### Summary of Angles
1. \( m\angle ABC = 60^\circ \)
2. \( m\angle BCA = 60^\circ \)
3. \( m\angle CAB = 60^\circ \)
4. \( m\angle BCD = 60^\circ \)
5. \( m\angle BDC = 60^\circ \)
6. \( m\angle CBD = 60^\circ \)
7. \( m\angle EDC = 60^\circ \)
8. \( m\angle ECD = 60^\circ \)
9. \( m\angle CED = 60^\circ \)
10. \( m\angle ECF = 60^\circ \)
11. \( m\angle EFC = 60^\circ \)
12. \( m\angle CEF = 60^\circ \)
All of the angles found in triangles \( ABC \), \( CDE \), and \( ECF \) are \( 60^\circ \) due to the properties of equilateral triangles and the given conditions of congruence.
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