Question

find ( moverarc{ad} ).
( moverarc{ad} = square^circ )

Explanation:

Step1: Recall cyclic quadrilateral theorem

In a cyclic quadrilateral, opposite angles sum to $180^\circ$.

Step2: Calculate $\angle B$

$\angle B = 180^\circ - 78^\circ = 102^\circ$

Step3: Sum of all arcs = $360^\circ$

Let $m\overset{\frown}{AD}=x$, $m\overset{\frown}{BC}=100^\circ$ (arc opposite $\angle D$), $m\overset{\frown}{CD}=73^\circ$, $m\overset{\frown}{AB}=180^\circ - 78^\circ$? No, use inscribed angle: arc = 2×inscribed angle opposite. Wait, correct: arc opposite $\angle A$ is $\overset{\frown}{BCD}=m\overset{\frown}{BC}+m\overset{\frown}{CD}$, so $\angle A=\frac{1}{2}(m\overset{\frown}{BCD})$, so $100^\circ=\frac{1}{2}(m\overset{\frown}{BC}+73^\circ)$ → $m\overset{\frown}{BC}=200^\circ-73^\circ=127^\circ$.

Step4: Calculate remaining arc $x$

$x + 127^\circ + 73^\circ + m\overset{\frown}{AB}=360^\circ$. $\angle D=\frac{1}{2}(m\overset{\frown}{AB}+m\overset{\frown}{BC})$ → $78^\circ=\frac{1}{2}(m\overset{\frown}{AB}+127^\circ)$ → $m\overset{\frown}{AB}=156^\circ-127^\circ=29^\circ$. Now $x=360^\circ-127^\circ-73^\circ-29^\circ$

Step5: Compute final value

$x=360-(127+73+29)=360-229=131^\circ$

Answer:

$131$