Question

triangle qst is isosceles, and rt bisects ∠t. what is true about ∠qrt? select two options. m∠qrt = 90° m∠qrt = m∠srt ∠qrt ≅ ∠stq m∠qrt = 2*m∠rtq ∠qrt ≅ ∠rtq

Explanation:

1. Analyze Triangle QST: Since \( \triangle QST \) is isosceles and \( \overline{RT} \) bisects \( \angle T \), by the properties of isosceles triangles and angle bisectors:

• \( \overline{RT} \) bisecting \( \angle T \) implies \( \angle QRT=\angle SRT \) (angle bisector definition), so \( m\angle QRT = m\angle SRT \) is true.
• In an isosceles triangle, the angle bisector from the vertex angle also acts as an altitude and median. So \( \angle QRT = 90^\circ \) (since it is a right angle in the isosceles triangle's angle - bisector/altitude/median case).
• For \( \angle QRT\cong\angle STQ \): There is no reason to assume these angles are congruent. The angle \( \angle STQ \) is part of the vertex angle of the isosceles triangle, and \( \angle QRT \) is a right angle (or related to the bisected angle), so this is false.
• For \( m\angle QRT = 2\times m\angle RTQ \): Let \( m\angle RTQ=x \), then \( m\angle QRT + m\angle RTQ+ m\angle RQT=180^\circ \). If \( \angle QRT = 90^\circ \), then \( 90^\circ+x + m\angle RQT = 180^\circ \), so \( m\angle RQT=90^\circ - x \), and there is no relation \( m\angle QRT = 2m\angle RTQ \) (since \( 90^\circ

eq2x \) in general for this triangle).

• For \( \angle QRT\cong\angle RTQ \): Since \( \angle QRT = 90^\circ \) (in the isosceles triangle with angle bisector) and \( \angle RTQ \) is an acute angle (part of the bisected angle), they are not congruent.

Answer:

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