Prove that in any triangle,the outer angle at a vertex is equal to the sum of the inner angles at the other two vertices.

let a triangle be GUN and the the outer angle on the vertex be angle X
now G+U+N = 180
and angle N+X = 180        (linear pair)
now, G+U+N = N + X
now here from both LHS and RHS the angle N will cut off
so, now we were remain of G+U = X    ( here G and U is your two interior angle and angle X is your outer angle)
here we proved this
 
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let there be a triangle ABC and an exterior angle at point A.let the exterior angle at a be X. here the points BC are in the opposite of angle X.
 in the triangle the measure of angle X can be found as: angle A+X=180 degree (linear pair)
                                                                         therefore angle X = 180-A
                                                                       so we have found the measure of angle X.
now we add the measures of angles : B+C
we will see that the measures of  angles B+C will be equal to measure of angle X.
This is a property of triangle. the result will be same( measure of vertex angle= sum of the interior angles st opposite side.) in any measurements taken.
hence we prove this statement.
 
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let a triangle be GUN and the the outer angle on the vertex be angle X
now G+U+N = 180
and angle N+X = 180? ? ? ? (linear pair)
now, G+U+N = N + X
now here from both LHS and RHS the angle N will cut off
so, now we were remain of G+U = X? ? ( here G and U is your two interior angle and angle X is your outer angle)
here we proved this
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