see the attached figure to better understand the problem
we know that
the sum of the internal angles of a triangle is equal to \(180\ degrees\)
so
\(m1+m2+m3=180\) -----> equation \(1\)
\(m3+m4=180\) ------> by suplementary angles
solve for m3
\(m3=180-m4\) -----> equation \(2\)
substitute equation \(2\) in equation \(1\)
\(m1+m2+(180-m4)=180\)
\(m1+m2+(180-m4)=180\)
\(m1+m2=m4\) ----> equation \(3\)
let's analyze each case
\(case\ a)\ m1 + m2 = m4\)
see the equation \(3\)
\(m1+m2=m4\)
\(This\ case\ is\ true\)
\(case\ b)\ m1 + m2 = m3\)
see the equation \(1\)
\(m1+m2+m3=180\\ m1+m2=180-m3\)
\(This\ case\ is\ not\ true\)
\(because\ (m1+m2)\ is\ equal\ to\ the\ suplementary\ angle\ of\ m3\)
\(case\ c)\ m4\ is\ greater\ than\ m2\)
see the equation \(3\)
\(m1+m2=m4\)
\(m4 > m2\)
so
\(This\ case\ is\ true\)
\(case\ d)\ m4\ is\ greater\ than\ m1\)
see the equation \(3\)
\(m1+m2=m4\)
\(m4 > m1\)
so
\(This\ case\ is\ true\)
\(case\ e)\ m3\ is\ greater\ than\ m2\)
see the equation \(1\)
\(m1+m2+m3=180\)
so
\(Not\ necessarily\ m3\ is\ greater\ than\ m2\)
\(case\ e)\ m2+m3=m4\)
see the equation \(3\)
\(m1+m2=m4\)
when m1=m3--------> m3+m2=m4
so
\(Not\ necessarily\)
therefore
the answer is
\(A.\ m1 + m2 = m4\)
\(C.\ m4\ is\ greater\ than\ m2\)
\(D.\ m4\ is\ greater\ than\ m1\)
