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prove: ( p x = p z = p y ) statements 1. ( overline { p r } ) bisects (…

Question

prove: ( p x = p z = p y )

statements

  1. ( overline { p r } ) bisects ( angle t r s )
  2. ( p ) is the incenter of ( \triangle r s t ) and lies on the angle bisector of ( angle t r s )
  3. ( p x = p z )
  4. ( overline { p s } ) bisects ( angle r s t )
  5. ( p ) lies on the angle bisector of ( angle r s t )

6.

  1. ( p x = p z = p y )

reasons

  1. given
  2. def. of incenter
  3. ( angle ) bisector thrm.
  4. given

5.
6.
7.

a. ( angle ) bisector theorem
b. trans. prop. of ( =)
c. def. of point
d. ( p z = p y )
e. def. of incenter
f. symmetric prop. of ( =)
g. perpendicular bisector theorem

Explanation:

Step1: Determine the reason for statement 5

Since \(P\) is the in - center of \(\triangle RST\) (from statement 2, def. of in - center), and \(\overline{PS}\) bisects \(\angle RST\) (statement 4, given), by the def. of in - center (which is the intersection of angle bisectors), \(P\) lies on the angle bisector of \(\angle RST\). So the reason for statement 5 is "Def. of in - center".

Step2: Determine statement 6

Since \(P\) lies on the angle bisector of \(\angle RST\) (statement 5), by the Angle Bisector Theorem (\(\angle\) Bisector Thrm.), the distances from \(P\) to the sides of the angle are equal. So \(PZ = PY\) (because \(PZ\) is the distance from \(P\) to \(RS\) and \(PY\) is the distance from \(P\) to \(ST\)). So statement 6 is \(PZ = PY\).

Step3: Determine the reason for statement 7

We know \(PX = PZ\) (statement 3) and \(PZ = PY\) (statement 6). By the Transitive Property of Equality (Trans. Prop. of \(=\)), if \(a = b\) and \(b = c\), then \(a = c\). Here \(a=PX\), \(b = PZ\), \(c = PY\). So the reason for statement 7 is "Trans. Prop. of \(=\)".

Answer:

  1. e. Def. of in - center
  2. d. \(PZ = PY\)
  3. b. Trans. Prop. of \(=\)