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(N/A) Let $\angle ABC = \beta$ and $\angle ACB = \gamma$. Since $\angle ABD$ is an exterior angle at vertex $B$,$\angle ABD = 180^\circ - \beta$. Sinc

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(N/A) Let $\angle ABC = \beta$ and $\angle ACB = \gamma$. Since $\angle ABD$ is an exterior angle at vertex $B$,$\angle ABD = 180^\circ - \beta$. Since $\angle ACE$ is an exterior angle at vertex $C$,$\angle ACE = 180^\circ - \gamma$. Given that $\angle ABD > \angle ACE$,we substitute the expressions: $180^\circ - \beta > 180^\circ - \gamma$. Subtracting $180^\circ$ from both sides,we get $-\beta > -\gamma$. Multiplying by $-1$ reverses the inequality sign: $\beta Since $\angle ACB (\gamma) > \angle ABC (\beta)$,the side opposite to the larger angle is longer. Therefore,the side opposite to $\angle ACB$ is $AB$ and the side opposite to $\angle ABC$ is $AC$. Thus,$AB > AC$ is incorrect based on the standard exterior angle theorem logic; if $\angle ABD > \angle ACE$,then $\angle ACB > \angle ABC$,which implies $AB > AC$ is false,rather $AC > AB$ is true.

$\angle ABD$ and $\angle ACE$ are exterior angles of $\Delta ABC$. If $\angle ABD > \angle ACE$,then prove that $AC > AB$.

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