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In the diagram above, the following relation holds true.

$mABC=mAB+mBC$

Start by drawing the radii $PA,$ $PB,$ and $PC,$ and by labeling the central angles corresponding to $AB,$ $BC,$ and $ABC.$

By definition, the arc measure is equal to the measure of the related central angle.$mABmBCmABC =m∠1=m∠2=m∠3 $

By the Angle Addition Postulate, $m∠3$ can be written as the sum of $m∠1$ and $m∠2.$
$mABC=m∠3⇕mABC=m∠1+m∠2 $

Finally, in the above formula, $mAB$ and $mBC$ can be substituted for $m∠1$ and $m∠2,$ respectively.
$mABC=m∠1+m∠2⇕mABC=mAB+mBC $