Question

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

Explanation:

1. For \( m\angle QRT = 90^\circ \): In an isosceles triangle \( QST \) with \( \overline{RT} \) bisecting \( \angle T \), by the property of isosceles triangles (angle - bisector, median, and altitude coincide in an isosceles triangle), if \( QS = QT \) (assuming the equal sides are \( QS \) and \( QT \)), then \( RT \) is also an altitude, so \( \angle QRT = 90^\circ \).
2. For \( m\angle QRT=m\angle SRT \): Since \( \overline{RT} \) bisects \( \angle T \), it divides \( \angle QTS \) into two equal angles. But also, in the isosceles triangle, the line \( RT \) is a median and altitude, and triangles \( QRT \) and \( SRT \) are congruent (by SAS, as \( QS = QT \), \( \angle QTR=\angle STR \), and \( RT = RT \)). So corresponding angles \( \angle QRT \) and \( \angle SRT \) are equal.
3. For \( \angle QRT\cong\angle STQ \): There is no geometric property that guarantees this congruence. The angles are not necessarily related in this way.
4. For \( m\angle QRT = 2\times m\angle RTQ \): In a right - angled triangle \( QRT \) (if \( \angle QRT = 90^\circ \)), the sum of angles is \( 180^\circ \). Let \( \angle RTQ=x \), then \( \angle RQT = 90 - x \). There is no reason for \( \angle QRT \) to be twice \( \angle RTQ \) in general.
5. For \( \angle QRT\cong\angle RTQ \): In triangle \( QRT \), if \( \angle QRT = 90^\circ \), then \( \angle RTQ\) and \( \angle RQT \) are acute angles and \( \angle RTQ+\angle RQT=90^\circ \), so \( \angle QRT\) (a right angle) cannot be congruent to \( \angle RTQ \) (an acute angle) unless the triangle is isosceles right - angled, but this is not guaranteed by the given information.

Answer:

A. \( m\angle QRT = 90^\circ \)
B. \( m\angle QRT=m\angle SRT \)