Question

given: ( ell ) is a perpendicular bisector of ( overline{ac} )
m is a perpendicular bisector of ( overline{bc} )
n is a perpendicular bisector of ( overline{ab} )
( ell ), m, and n intersect at p.
prove: ( ap = cp = bp )
statements

1. line ( ell ) is a perpendicular bisector of ( overline{ac} )
2. p is the circumcenter of ( \triangle abc ) and lies on the

perpendicular bisector of ( overline{ac} )

1. ( ap = cp )
2. m is a perpendicular bisector of ( overline{bc} )
3. p lies on the perpendicular bisector of ( overline{bc} )

6.

1. ( ap = cp = bp )

a. ( perp ) bis. thrm.
b. reflexive prop. of =
c. trans. prop. of =
d. def. circumcenter
e. def. perpendicular bisector

1. ( bp = cp )

Explanation:

Step1: Determine the reason for statement 5

Since \(P\) is the circum - center of \(\triangle ABC\) (from statement 2, which is based on the definition of the circum - center as the intersection of perpendicular bisectors), and \(m\) is a perpendicular bisector of \(\overline{BC}\) (statement 4, given), by the definition of the circum - center (\(d\)), \(P\) lies on the perpendicular bisector of \(\overline{BC}\).

Step2: Determine statement 6

Since \(P\) lies on the perpendicular bisector of \(\overline{BC}\), by the Perpendicular Bisector Theorem (\(\perp\) Bis. Thrm, \(a\)), \(BP = CP\).

Step3: Determine the reason for statement 7

We know \(AP = CP\) (statement 3) and \(BP = CP\) (statement 6). By the Transitive Property of equality (\(c\)) (\(AP = CP\) and \(BP = CP\) implies \(AP=CP = BP\)).

Answer:

1. \(d\) (Def. circumcenter)
2. \(BP = CP\) (\(a\) \(\perp\) Bis. Thrm)
3. \(c\) (Trans. Prop. of \(=\))