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

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Help would be much appreciated!

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up vote 3 down vote accepted

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 $d$.

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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\ $


$ \angle1 > \angle4\ $

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