Question

point z is equidistant from the vertices of δtuv. which must be true?
○ (overline{ta}congoverline{tb})
○ (overline{az}congoverline{bz})
○ (angle btzcongangle buz)
○ (angle tzacongangle tzb)

Explanation:

Step1: Recall the property of circum - center

A point that is equidistant from the vertices of a triangle is the circum - center of the triangle. Here, point Z is the circum - center of \(\triangle TUV\).

Step2: Analyze each option

• Option 1: \(\overline{TA}\cong\overline{TB}\): There is no information given to suggest that A and B are mid - points of equal segments of \(TU\) related to the circum - center property.
• Option 2: \(\overline{AZ}\cong\overline{BZ}\): There is no basis for this congruence from the fact that Z is equidistant from the vertices of \(\triangle TUV\).
• Option 3: \(\angle BTZ\cong\angle BUZ\): There is no reason for these angles to be congruent just because Z is the circum - center.
• Option 4: Since Z is the circum - center of \(\triangle TUV\), and \(ZA\perp TV\), \(ZB\perp TU\), and the circum - center is equidistant from the vertices. In right - triangles \(\triangle TZA\) and \(\triangle TZB\), \(TZ = TZ\) (common side) and \(ZA=ZB\) (Z is the circum - center, distances from circum - center to the mid - points of the perpendiculars from circum - center to the sides are equal in a sense related to the circle centered at Z passing through T, U, V). By the Hypotenuse - Leg congruence criterion in right - triangles, \(\triangle TZA\cong\triangle TZB\), so \(\angle TZA\cong\angle TZB\).

Answer:

\(\angle TZA\cong\angle TZB\)