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