Question

point z is equidistant from the vertices of \\( \triangle tuv \\). which must be true? \\( \overline{ta} \cong \overline{tb} \\) \\( \overline{az} \cong \overline{bz} \\) \\( \angle btz \cong \angle buz \\) \\( \angle tza \cong \angle tzb \\)

Explanation:

Step1: Recall the circumcenter property

The point equidistant from the vertices of a triangle is the circumcenter, which lies at the intersection of the perpendicular bisectors. So, \( Z \) is on the perpendicular bisectors of the sides, meaning \( ZA \perp TV \), \( ZB \perp TU \), and \( ZT = ZU = ZV \).

Step2: Analyze each option

• Option 1: \( \overline{TA} \cong \overline{TB} \): No info to say \( TA = TB \), as \( A,B \) are on different sides. Eliminate.
• Option 2: \( \overline{AZ} \cong \overline{BZ} \): \( AZ \) and \( BZ \) are on different perpendicular bisectors; no reason they're equal. Eliminate.
• Option 3: \( \angle BTZ \cong \angle BUZ \): In \( \triangle BTZ \) and \( \triangle BUZ \), \( ZT = ZU \) (circumradius), \( ZB \perp TU \) (so \( \angle ZBT = \angle ZBU = 90^\circ \)), and \( ZB \) is common. By HL, \( \triangle BTZ \cong \triangle BUZ \), so \( \angle BTZ \cong \angle BUZ \). This holds.
• Option 4: \( \angle TZA \cong \angle TZB \): \( \angle TZA \) is at \( TV \)'s bisector, \( \angle TZB \) at \( TU \)'s; no reason they're equal. Eliminate.

Answer:

\( \angle BTZ \cong \angle BUZ \) (the third option)