Write True or False and justify your answer in each of the following:
In the figure,if $AOB$ is a diameter and $\angle ADC = 120^{\circ}$,then $\angle CAB = 30^{\circ}$.
In the figure,if $AOB$ is a diameter and $\angle ADC = 120^{\circ}$,then $\angle CAB = 30^{\circ}$.
(TRUE) $AOB$ is a diameter of the circle with center $O$.
Consider the cyclic quadrilateral $ADCB$ (or $ADCEB$ as shown in the figure,but focusing on the cyclic quadrilateral $ADCB$ is sufficient).
Since $ADCB$ is a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle ADC + \angle ABC = 180^{\circ}$.
Given $\angle ADC = 120^{\circ}$,we have $120^{\circ} + \angle ABC = 180^{\circ}$.
$\angle ABC = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
In $\Delta ABC$,$\angle ACB = 90^{\circ}$ (Angle in a semicircle).
In $\Delta ABC$,the sum of angles is $180^{\circ}$.
$\angle CAB + \angle ABC + \angle ACB = 180^{\circ}$.
$\angle CAB + 60^{\circ} + 90^{\circ} = 180^{\circ}$.
$\angle CAB + 150^{\circ} = 180^{\circ}$.
$\angle CAB = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
Hence,the given statement is True.
Consider the cyclic quadrilateral $ADCB$ (or $ADCEB$ as shown in the figure,but focusing on the cyclic quadrilateral $ADCB$ is sufficient).
Since $ADCB$ is a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
Therefore,$\angle ADC + \angle ABC = 180^{\circ}$.
Given $\angle ADC = 120^{\circ}$,we have $120^{\circ} + \angle ABC = 180^{\circ}$.
$\angle ABC = 180^{\circ} - 120^{\circ} = 60^{\circ}$.
In $\Delta ABC$,$\angle ACB = 90^{\circ}$ (Angle in a semicircle).
In $\Delta ABC$,the sum of angles is $180^{\circ}$.
$\angle CAB + \angle ABC + \angle ACB = 180^{\circ}$.
$\angle CAB + 60^{\circ} + 90^{\circ} = 180^{\circ}$.
$\angle CAB + 150^{\circ} = 180^{\circ}$.
$\angle CAB = 180^{\circ} - 150^{\circ} = 30^{\circ}$.
Hence,the given statement is True.
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