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(B) The exterior angle of a triangle is equal to the sum of the two interior opposite angles.For $\Delta ABC$,the exterior angle $\angle ABD = \angle

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(B) The exterior angle of a triangle is equal to the sum of the two interior opposite angles.
For $\Delta ABC$,the exterior angle $\angle ABD = \angle BAC + \angle ACB = 110^{\circ}$.
The exterior angle $\angle ACE = \angle BAC + \angle ABC = 130^{\circ}$.
Since $\angle ACE > \angle ABD$,we have $(\angle BAC + \angle ABC) > (\angle BAC + \angle ACB)$,which implies $\angle ABC > \angle ACB$.
In a triangle,the side opposite to the larger angle is longer. Therefore,the side opposite to $\angle ABC$ (which is $AC$) must be greater than the side opposite to $\angle ACB$ (which is $AB$).
Thus,$AC > AB$,which means $AB < AC$.
Therefore,the statement $AB > AC$ is False.

State whether the following statement is true or false:
$(1)$ $\angle ABD$ and $\angle ACE$ are the exterior angles of $\Delta ABC$. If $\angle ABD = 110^{\circ}$ and $\angle ACE = 130^{\circ}$,then $AB > AC$.

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State whether the following statement is true or false:$(1)$ $\angle ABD$ and $\angle ACE$ are the exterior angles of $\Delta ABC$. If $\angle ABD = 110^{\circ}$ and $\angle ACE = 130^{\circ}$,then $AB > AC$.