Let's verify each case to determine the solution to the problem.
Statements
case A) m∠3 + m∠4 = [tex]180\°[/tex]
The statement is True
Because
Angle 3 and Angle 4 are supplementary angles
case B) m∠2 + m∠4 + m∠6 = [tex]180\°[/tex]
The statement is True
Because
The sum of the internal angles of a triangle is always equal to [tex]180\°[/tex]
case C) m∠2 + m∠4 = m∠5
The statement is True
Because
we know that
m∠2 + m∠4 + m∠6 = [tex]180\°[/tex] -----> equation A (see case B)
m∠5 + m∠6 = [tex]180\°[/tex] -------> by supplementary angles
m∠6 = [tex]180\°[/tex]-m∠5 -------> equation B
substitute equation B in equation A
m∠2 + m∠4 + [tex]180\°[/tex]-m∠5 = [tex]180\°[/tex]
m∠2 + m∠4 = m∠5 ---------> is ok
case D) m∠1 + m∠2= [tex]90\°[/tex]
The statement is False
Because
m∠1 + m∠2= [tex]180\°[/tex] -------> by supplementary angles
case E) m∠4 + m∠6=m∠2
The statement is False
Because
we know that
m∠2 + m∠4 + m∠6 = [tex]180\°[/tex]
m∠4 + m∠6 = [tex]180\°[/tex]-m∠2
m∠4 + m∠6=m∠2
[tex]180\°[/tex]-m∠2= m∠2
[tex]180\°[/tex]=2m∠2
m∠2=[tex]90\°[/tex]
the statement only will be true when the triangle be right triangle and
m∠2=[tex]90\°[/tex]
case F) m∠2 + m∠6 = m∠5
The statement is False
Because
we know that
m∠5 + m∠6 = [tex]180\°[/tex] -------> by supplementary angles
m∠5= [tex]180\°[/tex]-m∠6 -----> equation A
m∠2 + m∠6 = m∠5 --------> equation B (given equation)
substitute equation A in equation B
m∠2 + m∠6 = [tex]180\°[/tex]-m∠6
m∠2 + 2m∠6 = [tex]180\°[/tex]
the statement only will be true when the triangle be isosceles and
m∠4=m∠6
therefore
the answer is
m∠3 + m∠4 = [tex]180\°[/tex]
m∠2 + m∠4 + m∠6 = [tex]180\°[/tex]
m∠2 + m∠4 = m∠5
