Question

given: \\(\overline{pd}\\) is the perpendicular bisector of \\(\overline{ab}), and point \\(h\\) is on \\(\overline{pd}). prove: the points on the perpendicular bisector of a line segment are equidistant from the segment endpoints. move options to the table to complete the proof.

statement	reason
__ = __	reflexive property
m\\(\angle\\)__ = m\\(\angle\\)__ = 90°	definition of perpendicular lines
\\(\triangle\\)__ \\(\cong\\) \\(\triangle\\)__	sas
\\(\overline{ha}\\) \\(\cong\\) \\(\overline{hb}\\)	cpctc

Explanation:

Step1: Define bisector

$AD = BD$

Step2: Apply reflexive property

$HD=HD$

Step3: Define perpendicular lines

$m\angle ADH=m\angle BDH = 90^{\circ}$

Step4: Prove triangle - congruence

$\triangle ADH\cong\triangle BDH$ (by SAS since $AD = BD$, $\angle ADH=\angle BDH$, and $HD = HD$)

Step5: Use CPCTC

Since $\triangle ADH\cong\triangle BDH$, by CPCTC (Corresponding Parts of Congruent Triangles are Congruent), $HA\cong HB$

Answer:

Statement 1: $AD = BD$; Reason 1: Definition of bisector
Statement 2: $HD = HD$; Reason 2: Reflexive property
Statement 3: $m\angle ADH=m\angle BDH = 90^{\circ}$; Reason 3: Definition of perpendicular lines
Statement 4: $\triangle ADH\cong\triangle BDH$; Reason 4: SAS
Statement 5: $HA\cong HB$; Reason 5: CPCTC