Triangle congruence proofs and angle calculations

 If  $\overline{DA}$  bisects  $\overline{IE}$  IE, then

•  $\overline{ID}\ \cong\ \overline{EA}$  

•  $\overline{IA}\ \cong\ \overline{EA}$  
  

• <IDA  $\cong$  <EDA

• <IAD and <EAD are right angles

Tags

CCSS.HSG.CO.C.9

If L is the midpoint of  $\overline{TJ}$ , what must be true?

•  $\overline{TL}\ \cong\ \overline{JT}$  

•  $\overline{JL}\ \cong\ \overline{TJ}$ 
  

•  $\overline{TL}\ \cong\ \overline{LJ}$  

• T is also the midpoint of  $\overline{JL}$  

Tags

CCSS.HSG.CO.C.10

If  $\overline{AD}$ bisects <EDI, then

• <ADE=<IDA

•  $\overline{IA}\ \cong\ \overline{AE}$ 
  

•  $\overline{ID}\ \cong\ \overline{ED}$ 
  

• All of the above

Tags

CCSS.HSG.CO.C.9

 If  $\overline{DA}$  is perpendicular to  $\overline{IE}$ , then we can immediately conclude that

•  $\overline{IA}\ \cong\overline{AE}$  

• <IAD $\cong$  <DAE

• <IDA $\cong$ <EDA

• <IAD and <DAE are right angles

Tags

CCSS.HSG.CO.C.9

If  $\overline{DA}$  is the perpendicular bisector of  $\overline{EI}$ , then which is NOT true.

• <IAD and <DAE are right angles

•  $\overline{DI}\ \cong\overline{DE}$  

•  $\overline{EA}\ \cong\overline{IA}$ 
  

• Both 1 and 3 are correct

Tags

CCSS.HSG.CO.C.9

An angle bisector cuts an angle into two congruent _________

• Angles

• Segments

• Mysteries

Tags

CCSS.HSG.CO.C.9

A segment bisector cuts a segment into two congruent

• Segments

• Angles

• Mysteries

Tags

CCSS.HSG.CO.A.1

Fill in the missing proofs.

• Transitive Property, SAS(Side-Angle-Side)
• Reflexive Property, SAS(Side-Angle-Side)
• Reflexive Property, Vertical Angles Thm.
• Transitive Property, SSS(Side-Side-Side)

Tags

CCSS.HSG.SRT.B.5

What is the "statement" for step 2 of the proof?

• $\overline{AD}\ \cong\overline{AD}$
  

• $\overline{AD}\ \cong\overline{DA}$

• $\overline{HD}\ \cong\overline{DN}$

• $\overline{HA}\ \cong\overline{AN}$
  

What is the "statement" for step 3 of the proof?

• $\overline{AD}\ \cong\overline{DA}$

• $\overline{AN}\ \cong\overline{AH}$

• $\overline{HD}\ \cong\overline{ND}$

• $\overline{HD}\ \cong\overline{DN}$