Question

1. write a paragraph proof for the following conjecture

given: \\( \overline { q s } \\) bisects \\( \angle p q r \\)
\\( m \angle p q s = 4 5 ^ { \circ } \\)
prove: \\( \square p q r \\) is a right triangle
answer:

Explanation:

Step1: Use the angle - bisector definition

Since \( \overline{QS} \) bisects \( \angle PQR \), by the definition of an angle bisector, \( \angle PQS=\angle SQR \). Given that \( m\angle PQS = 45^{\circ} \), then \( m\angle SQR=45^{\circ} \).

Step2: Calculate \( m\angle PQR \)

Using the angle - addition postulate \( m\angle PQR=m\angle PQS + m\angle SQR \). Substitute \( m\angle PQS = 45^{\circ} \) and \( m\angle SQR = 45^{\circ} \) into the formula: \( m\angle PQR=45^{\circ}+45^{\circ}=90^{\circ} \).

Step3: Use the right - triangle definition

A triangle is a right triangle if one of its angles is a right angle (\( 90^{\circ} \)). Since \( \angle PQR = 90^{\circ} \) in \( \triangle PQR \), by the definition of a right triangle, \( \triangle PQR \) is a right triangle.

Answer:

Since \( \overline{QS} \) bisects \( \angle PQR \), \( \angle PQS=\angle SQR \). Given \( m\angle PQS = 45^{\circ} \), then \( m\angle SQR = 45^{\circ} \). By the angle - addition postulate \( m\angle PQR=m\angle PQS + m\angle SQR=45^{\circ}+45^{\circ}=90^{\circ} \). A triangle is a right triangle if one of its angles is \( 90^{\circ} \). So, \( \triangle PQR \) is a right triangle.