Question

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5 - 2 reteach to build understanding
envision geometry
savvasrealize.com
bisectors in triangles

1. write the letter of each figure beside its definition.

the circumcenter is the point of concurrency of the perpendicular bisectors of a
triangle.
the circumscribed circle is centered at the circumcenter and contains the
vertices of a triangle.
the incenter is the point of concurrency of the angle bisectors of a triangle.
the inscribed circle is centered at the incenter, and the sides of the triangle
are tangent to the circle.

1. timothy says ( px = py = pz ).

a. what mistake did timothy likely make?
b. from what points is ( p ) equidistant? explain your
reasoning.

1. complete the sentences below to find the value of ( x ).

( overline { ap } , overline { bp } ), and ( overline { cp } ) are the ____ of ( \triangle abc ).
so ( p ) is the ____ of ( \triangle abc ).
the incenter is equidistant from the ____
of ( \triangle abc ), so ( ps = )____.
therefore, ( x = )____.

Explanation:

Step1: Solve 1

• For the circum - center (point of concurrency of perpendicular bisectors) and circum - scribed circle (centered at circum - center and contains vertices), we look for the figure with perpendicular bisectors and the circle passing through vertices. Figure A shows the perpendicular bisectors and the circum - scribed circle.
• For the in - center (point of concurrency of angle bisectors) and in - scribed circle (centered at in - center and sides are tangent to the circle), Figure D shows the angle bisectors and the in - scribed circle. Figure B shows angle bisectors (in - center concept related) and Figure C shows perpendicular bisectors (circum - center concept related).
• The circum - center and circum - scribed circle: A
• The in - center: B
• The in - scribed circle: D

Step2: Solve 2a

• Timothy likely confused the properties of the circum - center and in - center. The circum - center is equidistant from the vertices of the triangle, while the in - center is equidistant from the sides of the triangle. P is the in - center (from the figure, since we have perpendiculars to the sides, which are the distances from the in - center to the sides). So, he confused the in - center (distance to sides) with the circum - center (distance to vertices).

Step3: Solve 2b

• Since P is the in - center (point of concurrency of angle bisectors), by the property of the in - center, it is equidistant from the sides of the triangle. The lengths \(PX\), \(PY\), and \(PZ\) represent the distances from the in - center \(P\) to the sides \(AB\), \(AC\), and \(BC\) respectively.

Step4: Solve 3

• \(\overline{AP}\), \(\overline{BP}\), and \(\overline{CP}\) are the angle bisectors of \(\triangle ABC\).
• So \(P\) is the in - center of \(\triangle ABC\).
• The in - center is equidistant from the sides of \(\triangle ABC\), so \(PS = PT\) (because \(PT\) and \(PS\) are the distances from the in - center to the sides).
• Since \(PT = 9\), therefore \(x=9\).

Answer:

1.

• The circum - center and circum - scribed circle: A
• The in - center: B
• The in - scribed circle: D

2.

• a. He confused the in - center (distance to sides) with the circum - center (distance to vertices).
• b. \(P\) is equidistant from the sides of \(\triangle ABC\) because \(P\) is the in - center (property: in - center is equidistant from the sides of a triangle).

3.

• \(\overline{AP}\), \(\overline{BP}\), and \(\overline{CP}\) are the angle bisectors of \(\triangle ABC\).
• So \(P\) is the in - center of \(\triangle ABC\).
• The in - center is equidistant from the sides of \(\triangle ABC\), so \(PS = PT\).
• Therefore, \(x = 9\).