Question

triangle qst is isosceles, and rt bisects \\( \angle t \\).
what is true about \\( \angle qrt \\)? choose two correct answers.
\\( m\angle qrt=m\angle srt \\)
\\( \angle qrt\cong\angle stq \\)
\\( m\angle qrt = 2\cdot m\angle rtq \\)
\\( \angle qrt\cong\angle rtq \\)
\\( m\angle qrt = 90 ^ { \circ } \\)

Explanation:

Step1: Use the property of angle - bisector

Since \(RT\) bisects \(\angle T\), then \(\angle QTR=\angle STR\). Also, in \(\triangle SRT\) and \(\triangle QRT\), \(ST = QT\) (isosceles triangle \(QST\)), \(RT = RT\) (common side), and \(\angle QTR=\angle STR\). By \(SAS\) (Side - Angle - Side) congruence criterion, \(\triangle SRT\cong\triangle QRT\). So, \(m\angle QRT=m\angle SRT\).

Step2: Use the exterior - angle property of a triangle

Let \(\angle RTQ = x\). In \(\triangle QRT\), \(\angle QRT\) is an exterior angle of \(\triangle RTQ\) (if we consider a different perspective, or use the fact that in an isosceles triangle \(QST\) with \(RT\) as a bisector and using triangle angle - sum and congruence). The sum of the interior angles of a triangle: In \(\triangle QRT\), \(\angle Q+\angle QTR+\angle QRT = 180^{\circ}\), and in \(\triangle SRT\) (congruent to \(\triangle QRT\)). Also, using the property that \(\angle QRT\) and \(\angle RTQ\) relationship. Since \(\triangle SRT\cong\triangle QRT\), and considering the angle - sum in \(\triangle QRT\) and the fact that \(\angle QRT\) is related to \(\angle RTQ\). If we assume \(\angle RTQ=\alpha\), then \(\angle QRT = 90^{\circ}\) (because in an isosceles triangle \(QST\) with \(RT\) as the angle - bisector and using the property that \(SR = RQ\) (from congruence of \(\triangle SRT\) and \(\triangle QRT\)) and \(ST = QT\), the triangle \(QRT\) and \(SRT\) are congruent right - angled triangles. \(m\angle QRT = 90^{\circ}\)

Answer:

\(m\angle QRT=m\angle SRT\), \(m\angle QRT = 90^{\circ}\)