Question

Given: Point B is on the perpendicular bisector of AC¯¯¯¯¯. BD¯¯¯¯¯ bisects AC¯¯¯¯¯ at point D.

Prove: B is equidistant from A and C.

A 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?



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Statement Reason
Point B is on the perpendicular bisector of AC¯¯¯¯¯
. BD¯¯¯¯¯
bisects AC¯¯¯¯¯
at point D.

Given
AD¯¯¯¯¯≅CD¯¯¯¯¯
Definition of bisector
∠ADB
and ∠CDB
are right angles. Definition of perpendicular
Press enter key to drop response here.Response input area All right angles are congruent.
BD¯¯¯¯¯≅BD¯¯¯¯¯
Press enter key to drop response here.Response input area
△ADB≅△CDB
Press enter key to drop response here.Response input area
AB¯¯¯¯¯≅CB¯¯¯¯¯
​CPCTC
​
AB=CB
Definition of congruent segments
B is equidistant from A and C. Definition of equidistant

Answer 1

Missing entries:

- ∠ADB ≅ ∠CDB
- Reason for BD ≅ BD: Reflexive property (BD is congruent to itself)
- Reason for △ADB ≅ △CDB: SAS (Side-Angle-Side)