Angle Bisector and Center

The INCENTER is the intersection of which special segments?

• Angle Bisectors

• Perpendicular Bisectors

• Medians

• Altitudes

Tags

CCSS.HSG.C.A.3

The figure is an example of a(n) ...

• angle bisector

• perpendicular bisector

• median

• midsegment

The place where all three angle bisectors meet is called the ...

• Circumcenter

• Orthocenter

• Incenter

• Centroid

Tags

CCSS.HSG.C.A.3

What type of center is point M?

• Circumcenter

• Incenter

• Centroid

• Orthocenter

Tags

CCSS.HSG.C.A.3

What is the missing statement in the proof?

• 2m∠ABD = m∠ABD + m∠ABD

• m∠ABD = m∠DBC

• m∠ABC = m∠ABC

• m∠ABD +m∠ABD = m∠ABD + m∠DBC

Tags

CCSS.4.MD.C.7

If S is the incenter, what type of segments are drawn?

• medians

• angle bisectors

• perpendicular bisectors

• altitudes

Tags

CCSS.HSG.C.A.3

Find the m<WYZ.

• 9°

• 18°

• 36°

Tags

CCSS.HSG.CO.C.9

What is the missing statement in the proof?

• 2m∠ABD=m∠ABD+m∠ABD

• m∠ABD=m∠DBC

• m∠ABC=m∠ABC

• m∠ABD+m∠ABD=m∠ABD+m∠DBC

Tags

CCSS.4.MD.C.7

• m∠SQP=29, m∠PQR=58
• m∠SQP=29, m∠PQR=29
• m∠SQP=58, m∠PQR=58
• m∠SQP=58, m∠PQR=29

Tags

CCSS.7.G.B.5

Given: m∠LAN = 30⁰ , m∠1 = 15⁰
Prove: AM bisects ∠LAN
Step 1: m∠LAN = 30⁰ 
Step 2: m∠1 + m∠2 = m∠LAN
Step 3: m∠1 + m∠2 = 30⁰
Step 4: m∠1 = 15⁰
Step 5: 15⁰ + m∠2 = 30⁰
Step 6: m∠2 = 15⁰
Step 7: m∠1 = m∠2
What is the justification for Step 7?

• Subtraction property of equality

• Transitive property of equality

• Addition property of equality

• Definition of congruent angles

Tags

CCSS.HSG.SRT.B.5