Ambiguity in the property that opposite angles of a cyclic quadrilaterals are supplementary

Question

Consider a convex cyclic quadrilateral ABCD.

A basic property of one such quadrilateral is that $\measuredangle \, BCD + \measuredangle \, DAB = 180^{\circ}$.

One way to settle this equality is by resorting to the inscribed angle theorem: indeed, if $O$ is the center of the circumference passing through $A$, $B$, $C$, and $D$, then

$\measuredangle \, BCD + \measuredangle \, DAB = \frac{1}{2}\left(\measuredangle \, BOD + \measuredangle \, DOB\right)$.

The conclusion follows from the previous line because, if we measure the angles $\angle \, BOD$ and $\angle \, DOB$ in a proper way, we get that $\measuredangle \, BOD + \measuredangle \, DOB = 360^{\circ}$.

What's the convention about angle measurement that needs to be made in order to unambiguously ascertain that $\measuredangle \, BOD + \measuredangle \, DOB = 360^{\circ}$? The thing is that if one doesn't look at $\angle \, DOB$ from the right perspective (so to speak), one may end up saying $\measuredangle \, BOD=\measuredangle \, DOB$ (instead of $\measuredangle \, BOD + \measuredangle \, DOB = 360^{\circ}$).

Hope you don't find this question too näive for this site.

Thanks in advance for your comments, suggestions, and replies.

Answer 1

Your question involves the naming convention of an angle.

"An angle whose size is between $90^0$ and $180^0$ shoudl be prefixed with the word 'reflex'."

Refering the the figure attached.

$\alpha' = \angle BOD = \angle DOB$.

$\gamma'$ should be named as "reflex $\angle BOD$ or reflex $\angle DOB$.

Added: In geobegra, the naming convention of an angle is anti-cloclockwise. Thus, saying $\angle DOB$ yields $\alpha'$; while saying $\angle BOD$ yields $\gamma'$.

Answer 2

Now focus on the figure. I definitely agree with you that here $\angle BOD = \angle DOB $. You are 100% correct in that. Well I don't know about you but I have just gone through standard 9 and that time our teacher told that it is $\measuredangle \, BCD + \measuredangle \, DAB = \frac{1}{2}\left(\measuredangle \, BOD + \text{reflex} \measuredangle \, BOD\right)$. So this should be the formula because $ \angle BOD = \angle DOB$ is not $360$ unless we take one to be non reflex and other as reflex angle leaving the straight line case. Our teacher also told us that many times it is written $BOD+DOB$. In those places you have to assume $BOD$ to be non reflex angle an $DOB$ to be the reflex angle. Sometimes it might be opposite too.

You can also look at the proof here. Do remember that arcs plays a good amount of role in this proof.