Question

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

Explanation:

Step1: Recall inscribed angle theorem

An inscribed angle is half its intercepted arc.

Step2: Find arc $\widehat{BD}$

$\angle BAD = 56^\circ$, so $m\widehat{BD} = 2\times56^\circ = 112^\circ$

Step3: Find arc $\widehat{AD}$

$\angle ABD = 90^\circ$, so $m\widehat{AD} = 2\times90^\circ = 180^\circ$
Wait, correct: $\angle ABD$ intercepts $\widehat{AD}$, so $m\widehat{AD}=2\times90^\circ=180^\circ$. Given $m\widehat{AD}$'s visible arc is $137^\circ$, no—total circle is $360^\circ$. First find arc $\widehat{AB}$: $\angle ADB$? No, use quadrilateral inscribed angles sum. Wait, inscribed quadrilateral opposite angles sum to $180^\circ$. $\angle A + \angle C = 180^\circ$, $\angle B + \angle D = 180^\circ$. But better:
Total circle: $360^\circ = m\widehat{AB} + m\widehat{BC} + m\widehat{CD} + m\widehat{DA}$
$\angle BAD=56^\circ$ intercepts $\widehat{BCD}$, so $m\widehat{BCD}=2\times56^\circ=112^\circ$? No, $\angle BAD$ is inscribed over $\widehat{BD}$, so $m\widehat{BD}=112^\circ$. $m\widehat{BD}=m\widehat{BC}+m\widehat{CD}=112^\circ$
$\angle ABC=90^\circ$ intercepts $\widehat{ADC}$, so $m\widehat{ADC}=2\times90^\circ=180^\circ$. $m\widehat{ADC}=m\widehat{AD}+m\widehat{CD}=180^\circ$. Given $m\widehat{AD}=137^\circ$, so $m\widehat{CD}=180^\circ - 137^\circ=43^\circ$
Wait, verify: $m\widehat{BD}=m\widehat{BC}+43^\circ=112^\circ$, so $m\widehat{BC}=69^\circ$. Total arcs: $137+43+69+m\widehat{AB}=360$, $m\widehat{AB}=360-249=111^\circ$. $\angle BCD$ would be half $\widehat{ABD}=0.5\times(111+137)=124^\circ$, and $\angle A=56^\circ$, $56+124=180$, which fits inscribed quadrilateral. Correct.

Step2 (corrected): Use $\angle B$ for $\widehat{AD}$

$\angle ABC=90^\circ$ intercepts $\widehat{ADC}$, so $m\widehat{ADC}=2\times90^\circ=180^\circ$

Step3: Calculate $m\widehat{CD}$

$m\widehat{CD}=m\widehat{ADC}-m\widehat{AD}=180^\circ-137^\circ$

Answer:

$43^\circ$