Question

mr. sale & ms. thornhill
geometry proof name: soudar sagaralek
hw#15 band: f
problem 10

1. ∠bac ≅ ∠dca 1. given

2 2. given

1. 3.
2. △abc ≅ △cda 4.

problem 11

1. ∠f ≅ ∠i 1.
2. ∠ ≅ ∠ 2.
3. 3.
4. △efg ≅ △hij 4.

problem 12

1. ∠o ≅ ∠m 1. given
2. 2. given
3. ∠klo ≅ ∠ 3.
4. △klo ≅ △nlm 4.
5. ∠k ≅ ∠n 5. cpctc

problem 13

1. ∠p ≅ ∠r 1. given
2. 2. given
3. 3. reflexive
4. △pqs ≅ △rsq 4.

problem 14

1. overline{ac} || overline{bd} 1. given
2. 2. given
3. ∠cad ≅ ∠bda 3.
4. 4. reflexive property
5. △acd ≅ △____ 5.

Explanation:

Problem 10

Step1: Identify the second given

Since we are proving triangle congruence and we have one angle congruent (\(\angle BAC\cong\angle DCA\)) from statement 1. For \(ASA\) (Angle - Side - Angle) or \(AAS\) (Angle - Angle - Side) or \(SAS\) (Side - Angle - Side) congruence, we need another pair of sides or angles. Looking at the figure, assume \(AC = CA\) (common side).
Statement 2: \(AC=CA\)
Reason 2: Reflexive Property

Step2: Identify the third given

If we assume \(AB\parallel CD\) (from the figure's appearance, which would give \(\angle BCA\cong\angle DAC\) by alternate - interior angles).
Statement 3: \(\angle BCA\cong\angle DAC\)
Reason 3: Alternate - interior angles (if \(AB\parallel CD\))

Step3: Determine the congruence rule

Since we have \(\angle BAC\cong\angle DCA\), \(AC = CA\), and \(\angle BCA\cong\angle DAC\), by \(ASA\) (Angle - Side - Angle) congruence.
Reason 4: \(ASA\)

Problem 11

Step1: Identify the second pair of angles

For triangle congruence (\(\triangle EFG\) and \(\triangle HIJ\)), we know \(\angle F\cong\angle I\) (statement 1). If we assume \(\angle E\cong\angle H\) (by some given or figure property).
Statement 2: \(\angle E\cong\angle H\)
Reason 2: Given (or from figure property)

Step2: Identify the side

If \(FG = IJ\) (from the figure's marking)
Statement 3: \(FG = IJ\)
Reason 3: Given

Step3: Determine the congruence rule

Since we have \(\angle F\cong\angle I\), \(\angle E\cong\angle H\), and \(FG = IJ\), by \(AAS\) (Angle - Angle - Side) congruence.
Reason 4: \(AAS\)

Problem 12

Step1: Identify the second given

Assume \(OL = ML\) (from the figure's marking, as the segments are marked equal)
Statement 2: \(OL=ML\)
Reason 2: Given

Step2: Identify the third angle

Since \(\angle KLO\) and \(\angle NLM\) are vertical angles (from the figure's intersection of lines \(KN\) and \(OM\))
Statement 3: \(\angle KLO\cong\angle NLM\)
Reason 3: Vertical angles

Step4: Determine the congruence rule

Since \(\angle O\cong\angle M\), \(OL = ML\), and \(\angle KLO\cong\angle NLM\), by \(ASA\) (Angle - Side - Angle) congruence.
Reason 4: \(ASA\)

Problem 13

Step1: Identify the second given

Assume \(\angle PQS\cong\angle RSQ\) (from the figure's marking or given)
Statement 2: \(\angle PQS\cong\angle RSQ\)
Reason 2: Given

Step2: Identify the third given

Assume \(SQ = SQ\) (common side)
Statement 3: \(SQ=SQ\)
Reason 3: Reflexive Property

Step3: Determine the congruence rule

Since \(\angle P\cong\angle R\), \(\angle PQS\cong\angle RSQ\), and \(SQ = SQ\), by \(AAS\) (Angle - Angle - Side) congruence.
Reason 4: \(AAS\)

Problem 14

Step1: Identify the second given

Assume \(\angle ACD\cong\angle BDC\) (alternate - interior angles as \(AC\parallel BD\))
Statement 2: \(\angle ACD\cong\angle BDC\)
Reason 2: Alternate - interior angles

Step2: Identify the side

Assume \(CD = DC\) (common side)
Statement 4: \(CD = DC\)
Reason 4: Reflexive Property

Step3: Determine the triangle name

Since we have \(\angle CAD\cong\angle BDA\), \(\angle ACD\cong\angle BDC\), and \(CD = DC\), \(\triangle ACD\cong\triangle BDC\)
Statement 5: \(\triangle ACD\cong\triangle BDC\)
Reason 5: \(AAS\)

Answer:

Problem 10

• Statement 2: \(AC = CA\); Reason 2: Reflexive Property
• Statement 3: \(\angle BCA\cong\angle DAC\); Reason 3: Alternate - interior angles (if \(AB\parallel CD\))
• Reason 4: \(ASA\)

Problem 11

• Statement 2: \(\angle E\cong\angle H\); Reason 2: Given (or from figure property)
• Statement 3: \(FG = IJ\); Reason 3: Given
• Reason 4: \(AAS\)

Problem 12

• Statement 2: \(OL = ML\); Reason 2: Given
• Statement 3: \(\angle KLO\cong\angle NLM\); Reason 3: Vertical angles
• Reason 4: \(ASA\)

Problem 13

• Statement 2: \(\angle PQS\cong\angle RSQ\); Reason 2: Given
• Statement 3: \(SQ = SQ\); Reason 3: Reflexive Property
• Reason 4: \(AAS\)

Problem 14

• Statement 2: \(\angle ACD\cong\angle BDC\); Reason 2: Alternate - interior angles
• Statement 4: \(CD = DC\); Reason 4: Reflexive Property
• Statement 5: \(\triangle ACD\cong\triangle BDC\); Reason 5: \(AAS\)