Angle Bisector Theorem Practice

Find the missing length.

• 10

• 21

• 35

• 9

Tags

CCSS.HSG.CO.C.9

Solve for x.

• 5

• 10

• 3

• 8

Tags

CCSS.HSG.CO.C.9

Solve for x.

• 8

• 4

• 7

• 5

Tags

CCSS.HSG.CO.C.9

Solve for x.

• 6

• 8

• 9

• 10

Tags

CCSS.HSG.CO.C.9

If AD bisects <EDI, then

• <ADE=<IDA

• IA=AE

• ID=ED

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

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
What is the justification for Step 4?

• Given

• Substitution

• Angle addition postulate

• Definition of congruent angles

Tags

CCSS.HSG.SRT.B.5

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

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
What is the justification for Step 5?

• Given

• Substitution

• Angle addition postulate

• Definition of congruent angles

Tags

CCSS.HSG.SRT.B.5

If we are given segment RT bisects ∠VRS, what can we conclude. 

• ∠VRT≅∠SRT

• ∠RTV≅∠RTS

• ∠RVT≅∠RST

• VT≅∠TS

Tags

CCSS.HSG.CO.C.9