Question

which statements are always true regarding the diagram? check all that apply.

( m angle 3 + m angle 4 = 180 ^ { circ } )

( m angle 2 + m angle 4 + m angle 6 = 180 ^ { circ } )

( m angle 2 + m angle 4 = m angle 5 )

( m angle 1 + m angle 2 = 90 ^ { circ } )

( m angle 4 + m angle 6 = m angle 2 )

( m angle 2 + m angle 6 = m angle 5 )

Explanation:

Step1: Analyze \(m\angle3 + m\angle4=180^{\circ}\)

\(\angle3\) and \(\angle4\) are adjacent and form a linear - pair. By the linear - pair postulate, if two angles form a linear pair, then they are supplementary. So \(m\angle3 + m\angle4 = 180^{\circ}\) is always true.

Step2: Analyze \(m\angle2+m\angle4 + m\angle6=180^{\circ}\)

By the triangle - angle sum theorem, the sum of the interior angles of a triangle is \(180^{\circ}\). In the triangle with angles \(\angle2\), \(\angle4\), and \(\angle6\), \(m\angle2 + m\angle4+m\angle6=180^{\circ}\) is always true.

Step3: Analyze \(m\angle2 + m\angle4=m\angle5\)

By the exterior - angle theorem, an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. \(\angle5\) is an exterior angle of the triangle with interior angles \(\angle2\) and \(\angle4\). So \(m\angle2 + m\angle4=m\angle5\) is always true.

Step4: Analyze \(m\angle1 + m\angle2=90^{\circ}\)

There is no information in the diagram to suggest that \(\angle1\) and \(\angle2\) are complementary. They form a linear pair (\(m\angle1 + m\angle2 = 180^{\circ}\)), so \(m\angle1 + m\angle2=90^{\circ}\) is false.

Step5: Analyze \(m\angle4 + m\angle6=m\angle2\)

From the triangle - angle sum theorem \(m\angle2+m\angle4 + m\angle6=180^{\circ}\). If \(m\angle4 + m\angle6=m\angle2\), then \(2m\angle2=180^{\circ}\) (substitute \(m\angle4 + m\angle6\) with \(m\angle2\) in \(m\angle2+m\angle4 + m\angle6=180^{\circ}\)), which is only true when \(m\angle2 = 90^{\circ}\). But there is no general proof for this from the given diagram. So \(m\angle4 + m\angle6=m\angle2\) is false.

Step6: Analyze \(m\angle2 + m\angle6=m\angle5\)

Since \(m\angle2 + m\angle4=m\angle5\) (from the exterior - angle theorem) and \(m\angle2+m\angle4 + m\angle6=180^{\circ}\), if \(m\angle2 + m\angle6=m\angle5\), then \(2m\angle5=180^{\circ}\) (substitute \(m\angle2\) as \(m\angle5 - m\angle6\) into \(m\angle2+m\angle4 + m\angle6=180^{\circ}\), and \(m\angle4=m\angle5 - m\angle2=m\angle5-(m\angle5 - m\angle6)=m\angle6\)). There is no general proof for this from the given diagram. So \(m\angle2 + m\angle6=m\angle5\) is false.

Answer:

\(m\angle3 + m\angle4 = 180^{\circ}\), \(m\angle2 + m\angle4 + m\angle6=180^{\circ}\), \(m\angle2 + m\angle4=m\angle5\)