The external angle of a circle equals half the difference of its intercepted arcs. Please explain .?

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* This is a well known fact that an angle inscribed into a circle measures half the central angle subtending the same arc.

Inscribed angle=(1/2)*Central angle ==>

∠MBN=(∠MON)/2

∠AMB=(∠AOB)/2

** An exterior angle of a triangle is equal in measure to the sum of the two non-adjacent interior angles of the triangle.

Applying to the triangle BPM ==>

∠MBN=∠APB+∠AMB ==>

∠APB=∠MBN -∠AMB=(1/2)*(∠MON -∠AOB)

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RE:

The external angle of a circle equals half the difference of its intercepted arcs. Please explain .?

Assume PA M and PBN are secants to a circle from point P intersecting the circle at A ,M and B,N. I think we are required to prove angle APB is half the difference of the angle of intercepted arcs. I would like to draw a diagram but how? Please explain with proof or provide reference. Thanks a lot.

A. Always true.