Triangle angles How would I prove that, in any triangle, any of the exterior angles is bigger than any of the remote interior angles?

Help would be much appreciated!
 A: From your figure above, let
$m \angle1 = a$
$m \angle2 = b$
$m \angle3 = c$
$m \angle4 = d$
We must show that $a > c$ and $a > d$.
Since all of the angles of a triangle must sum to $180^\circ$, we have
$b + c + d = 180$
Also, since $\angle1$ and $\angle2$ are supplementary:
$a + b = 180$
Reversing the second equality and adding to the first, we have
$(b + c + d) + 180 = 180 + (a + b)$
$\therefore a = c + d$
$a$, $c$ and $d$ are all positive, so if $a = c + d$, then $a$ must be greater than both $c$ and $d$.
A: As $ \angle1 = \angle3 + \angle4 $  
The above is because of the rule "exterior angle of a triangle is equal to sum of two the interior opposite angles" .
as this is true 
1)   $ \angle1 = \angle3\ this\ is\ true\ iff \angle4=0 $   and
2)   $ \angle1 = \angle4\ this\ is\ true\ iff \angle3=0 $   
But the above two violates the definition that triangle has 3 angles .
so none of them is zero and they are also positive(as angles are always positive measuring anticlockwise by convention). so the above two possibilities of being equal are ruled out .
if a quantity is equal to sum of two other positive quantities.....
sum will be greater than either of the two quantities added .
so
$ \angle1 > \angle3\ $
and
$ \angle1 > \angle4\ $
A: I am going to do it in the simplest way so accept it
You know
1+2=180.........(A)
&     2+3+4=180........(B)
Putting (B) in(A) we get
1+2=2+3+4
So 1=3+4
Thus prooves 1 has be greater than 3 or 4
