133277
contestada

Given:
\overleftrightarrow{BD}
BD
B, D, with, \overleftrightarrow, on top is the perpendicular bisector of segment \overline{AC}
AC
start overline, A, C, end overline.
\overline{BD}
BD
start overline, B, D, end overline is 333 units long.
\overline{AC}
AC
start overline, A, C, end overline is 888 units long.
Naomi was asked to show that point DDD is equidistant from points AAA and CCC.

Respuesta :

A line bisector is a straight line that divides a given line into two equal parts. Thus the following steps are required by Naomi to show that point D is equidistant from points  A and C.

BD ⊥ AC (given)

BD = 3 units, and AC = 8 units.

BD is the perpendicular bisector of segment AC (given)

Thus,

<BDA  ≅ <BDC (right angles formed by a perpendicular bisector)

AD ≅ DC (equal parts of a bisected line)

AD ≅ DC = 4 units

Thus joining points B to A, and B to C,

BA ≅ BC.

So that applying Pythagoras theorem to ΔABD, we have:

[tex]/hyp/^{2}[/tex] = [tex]Adj 1^{2}[/tex] + [tex]Adj 2^{2}[/tex]

[tex]AB^{2}[/tex] = [tex]3^{2}[/tex] + [tex]4^{2}[/tex]

       = [tex]\sqrt{25}[/tex]

 AB      = 5 units

So that,

BA ≅ BC = 5 units

Therefore, it can be concluded that point D is equidistant from points A and C.

For more clarifications on a perpendicular bisector of a line, visit: https://brainly.com/question/929137

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