Question

find ( moverarc{ab} ), ( moverarc{bc} ), and ( moverarc{cd} ).

Explanation:

Step1: Recall inscribed angle theorem

The measure of an inscribed angle is half the measure of its intercepted arc. So, intercepted arc = 2 × inscribed angle.

Step2: Calculate $m\overset{\frown}{BC}$

$\angle D$ intercepts $\overset{\frown}{ABC}$, but $\angle A$ intercepts $\overset{\frown}{BCD}$. Wait, directly: $\angle DAB = 84^\circ$ intercepts $\overset{\frown}{BCD}$, which is $\overset{\frown}{BC}+\overset{\frown}{CD}$. We know $\overset{\frown}{AD}=98^\circ$, total circle is $360^\circ$. First, $\angle ABC=94^\circ$ intercepts $\overset{\frown}{ADC}=\overset{\frown}{AD}+\overset{\frown}{CD}$. So $\overset{\frown}{ADC}=2\times94^\circ=188^\circ$. Since $\overset{\frown}{AD}=98^\circ$, then $\overset{\frown}{CD}=188^\circ-98^\circ=90^\circ$.

Step3: Calculate $m\overset{\frown}{AB}$

$\angle ADC$ intercepts $\overset{\frown}{ABC}$. First, find $\angle ADC$: in cyclic quadrilateral, opposite angles sum to $180^\circ$, so $\angle ADC=180^\circ-94^\circ=86^\circ$. Then $\overset{\frown}{ABC}=2\times86^\circ=172^\circ$. We know $\overset{\frown}{CD}=90^\circ$, so $\overset{\frown}{AB}=360^\circ-\overset{\frown}{AD}-\overset{\frown}{CD}=360^\circ-98^\circ-90^\circ=172^\circ$? No, wait: $\angle BCD=180^\circ-84^\circ=96^\circ$, intercepts $\overset{\frown}{ABD}=\overset{\frown}{AB}+\overset{\frown}{AD}=2\times96^\circ=192^\circ$. So $\overset{\frown}{AB}=192^\circ-98^\circ=94^\circ$.

Step4: Calculate $m\overset{\frown}{BC}$

Total circle: $\overset{\frown}{AB}+\overset{\frown}{BC}+\overset{\frown}{CD}+\overset{\frown}{AD}=360^\circ$. So $\overset{\frown}{BC}=360^\circ-94^\circ-90^\circ-98^\circ=78^\circ$. Verify: $\angle DAB=84^\circ$ intercepts $\overset{\frown}{BC}+\overset{\frown}{CD}=78^\circ+90^\circ=168^\circ$, half is $84^\circ$, correct. $\angle ABC=94^\circ$ intercepts $\overset{\frown}{AD}+\overset{\frown}{CD}=98^\circ+90^\circ=188^\circ$, half is $94^\circ$, correct.

Answer:

$m\overset{\frown}{AB}=94^\circ$
$m\overset{\frown}{BC}=78^\circ$
$m\overset{\frown}{CD}=90^\circ$