Question

proving theorems about lines and angles: tutorial
m∠pea = m∠peb
ae = be
given
defining an arbitrary point
let p be any point on cd.
definition of ⊥ bisector (using angles)
pa = pb
transitive property of equality
cd is a ⊥ bisector of ab.
definition of ⊥ bisector (using midpoint)
δpea ≅ δpeb
(diagram)
reflexive property of equality
m∠pea = 90° m∠peb
= 90°
sas criterion for ≅ triangles
pe = pe
corresponding parts of ≅ triangles are ≅.
pa ≅ pb
definition of ≅

Explanation:

Step1: Match statements with reasons

• \(m\angle PEA = m\angle PEB\) is given.
• \(AE = BE\) is definition of \(\perp\) bisector (using midpoint).
• Let \(P\) be any point on \(\overleftrightarrow{CD}\) is defining an arbitrary point.
• \(PA = PB\) is Corresponding parts of \(\cong\) triangles are \(\cong\).
• \(\overleftrightarrow{CD}\) is a \(\perp\) bisector of \(\overline{AB}\) is definition of \(\perp\) bisector (using angles).
• \(\triangle PEA\cong\triangle PEB\) (diagram) is SAS criterion for \(\cong\) triangles.
• \(m\angle PEA = 90^{\circ}, m\angle PEB=90^{\circ}\) is Transitive Property of Equality.
• \(PE = PE\) is Reflexive Property of Equality.
• \(\overline{PA}\cong\overline{PB}\) is definition of \(\cong\).

Answer:

• \(m\angle PEA = m\angle PEB\): given
• \(AE = BE\): definition of \(\perp\) bisector (using midpoint)
• Let \(P\) be any point on \(\overleftrightarrow{CD}\): defining an arbitrary point
• \(PA = PB\): Corresponding parts of \(\cong\) triangles are \(\cong\)
• \(\overleftrightarrow{CD}\) is a \(\perp\) bisector of \(\overline{AB}\): definition of \(\perp\) bisector (using angles)
• \(\triangle PEA\cong\triangle PEB\) (diagram): SAS criterion for \(\cong\) triangles
• \(m\angle PEA = 90^{\circ}, m\angle PEB = 90^{\circ}\): Transitive Property of Equality
• \(PE = PE\): Reflexive Property of Equality
• \(\overline{PA}\cong\overline{PB}\): definition of \(\cong\)