Given: point b is on the perpendicular bisector of ac¯¯¯¯¯. bd¯¯¯¯¯ bisects ac¯¯¯¯¯ at point d. prove: b is equidistant from a and c. an isosceles triangle a b c. point d lies on side a c. an altitude is drawn from point b to point d. what are the missing parts that correctly complete the proof?

 Missing parts are,

In first blank,  ,

In second blank, SAS postulate

In third blank, CPCTC postulate

Step-by-step explanation:

Since, Here D is the mid point on the line segment AC.

And BD is a perpendicular to the line AC.

Therefore, In triangles ADB and CDB ( shown in figure)

AD\cong DC ( By the definition of mid point)

( right angles )

( reflexive)

Thus, By SAS ( side angle side )postulate,

So, by CPCTC( Corresponding parts of congruent triangles are congruent)

Now, By definition of congruent segment,

AB=CB

By definition of equidistant,

B is equally far from both A and C.

Point B is on the perpendicular bisector of AC. BD bisects Ac at point D.                  

                                                                 Given.

<ADB and <CDB are right angles          Definition of perpendicular.

<ADB = <CDB                                          All right angles are congruent.

BD = BD                                                    Reflexive property.

AB = CB                                                     Definition of congruent segments

B is equidistant from A and C                   Definition of equidistant.

i dont know

step-by-step explanation: