Question

consider δtuv in the figure below.
the perpendicular bisectors of its sides are \\(\overline{ad}\\), \\(\overline{bd}\\), and \\(\overline{cd}\\). they meet at a single point d.
(in other words, d is the circumcenter of δtuv.)
suppose \\(cd = 30\\), \\(uv = 142\\), and \\(ud = 78\\).
find \\(bv\\), \\(vd\\), and \\(tc\\).
note that the figure is not drawn to scale.

Explanation:

Step1: Find \( BV \)

Since \( \overline{BD} \) is the perpendicular bisector of \( \overline{UV} \), \( B \) is the midpoint of \( \overline{UV} \). So \( BV=\frac{1}{2}UV \). Given \( UV = 142 \), then \( BV=\frac{142}{2}=71 \).

Step2: Find \( VD \)

The circumcenter \( D \) is equidistant from all vertices of the triangle, so \( UD = VD \). Given \( UD = 78 \), then \( VD = 78 \).

Step3: Find \( TC \)

First, consider right triangle \( UCD \). We know \( UD = 78 \) and \( CD = 30 \). By the Pythagorean theorem, \( UC=\sqrt{UD^{2}-CD^{2}}=\sqrt{78^{2}-30^{2}}=\sqrt{(78 - 30)(78 + 30)}=\sqrt{48\times108}=\sqrt{5184}=72 \). Since \( \overline{CD} \) is the perpendicular bisector of \( \overline{TU} \), \( C \) is the midpoint of \( \overline{TU} \), so \( TC = UC = 72 \).

Answer:

\( BV = 71 \)
\( VD = 78 \)
\( TC = 72 \)