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pshichka [43]
3 years ago
12

HELP ASAP!!!!! Given: Point B is on the perpendicular bisector of AC¯¯¯¯¯. BD¯¯¯¯¯ bisects AC¯¯¯¯¯ at point D. Prove: B is equid

istant 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?

Mathematics
2 answers:
Harlamova29_29 [7]3 years ago
7 0

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.




AnnyKZ [126]3 years ago
6 0

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

                                                                 Given.


<h3>AD ≅ DC                                         Definition of bisector.</h3>

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


<ADB = <CDB                                          All right angles are congruent.


BD = BD                                                    Reflexive property.


<h3>Triangle △ADB ≅ △CDB               SAS congruence Postulate.</h3><h3>AB = CB                                            CPCTC </h3>

AB = CB                                                     Definition of congruent segments


B is equidistant from A and C                   Definition of equidistant.

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