Question

HELP ASAP!!!!! 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?

Answer 1

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

                                                                 Given.

AD ≅ DC                                         Definition of bisector.

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

<ADB = <CDB                                          All right angles are congruent.

BD = BD                                                    Reflexive property.

Triangle △ADB ≅ △CDB               SAS congruence Postulate.

AB = CB                                            CPCTC

AB = CB                                                     Definition of congruent segments

B is equidistant from A and C                   Definition of equidistant.

Answer 2

Answer:  Missing parts are,

In first blank,  $AD\cong DC$,

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)

$\angle BDA\cong \angle BDC$ ( right angles )

$BD\cong BD$ ( reflexive)

Thus, By SAS ( side angle side )postulate,

$\triangle ADB\cong \triangle CDB$

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

$AB\cong CB$

Now, By definition of congruent segment,

AB=CB

By definition of equidistant,

B is equally far from both A and C.