$D$ is a point on side $BC$ of $\Delta ABC$ such that $AD = AC$ (see figure). Show that $AB > AD$.
(N/A) In $\Delta DAC$,
$AD = AC$ (Given)
Therefore,$\angle ADC = \angle ACD$ (Angles opposite to equal sides are equal).
Now,$\angle ADC$ is an exterior angle for $\Delta ABD$.
By the exterior angle theorem,$\angle ADC > \angle ABD$.
Since $\angle ADC = \angle ACD$,we have $\angle ACD > \angle ABD$.
This implies $\angle ACB > \angle ABC$.
In $\Delta ABC$,the side opposite to the larger angle is longer. Therefore,$AB > AC$.
Since $AD = AC$,we conclude that $AB > AD$.
$AD = AC$ (Given)
Therefore,$\angle ADC = \angle ACD$ (Angles opposite to equal sides are equal).
Now,$\angle ADC$ is an exterior angle for $\Delta ABD$.
By the exterior angle theorem,$\angle ADC > \angle ABD$.
Since $\angle ADC = \angle ACD$,we have $\angle ACD > \angle ABD$.
This implies $\angle ACB > \angle ABC$.
In $\Delta ABC$,the side opposite to the larger angle is longer. Therefore,$AB > AC$.
Since $AD = AC$,we conclude that $AB > AD$.
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