Solution:

 OM \(\perp\) AB ... (Given)

∴ \(\text{AM} = \displaystyle \frac{1}{2} \times \text{AB}\) ... (Theorem)

∴ \(\text{AM} = \displaystyle \frac{1}{2} \times 16\)

∴ \(\text{AM} = 8\) ... (i)

Now, in right \(\triangle\)OMA,

 \(\text{OM}^2 + \text{AM}^2 = \text{OA}^2\) ... (Pythagoras’ Theorem)

∴ \(\text{OM}^2 + 8^2 = 10^2\)

∴ \(\text{OM}^2 + 64 = 100\)

∴ \(\text{OM}^2 = 100 - 64\)

∴ \(\text{OM}^2 = 36\)

∴ \(\text{OM} = \sqrt{36}\)

∴ \(\text{OM} = 6\) cm ... (ii)

Now, congruent chords of a circle are equidistant from the centre. ... (Theorem)

∴ ON = OM

∴ ON = 6 cm ... (iii)

The distance of the chords from the centre of the circle is 6 cm.

Solution:

 In right \(\triangle\)OMA,

 \(\text{OA}^2 = \text{OM}^2 + \text{AM}^2\) ... (Pythagoras’ Theorem)

∴ \(13^2 = 5^2 + \text{AM}^2\)

∴ \(169 = 25 + \text{AM}^2\)

∴ \(169 - 25 = \text{AM}^2\)

∴ \(144 = \text{AM}^2\)

i.e. \(\text{AM}^2 = 144\)

∴ \(\text{AM} = \sqrt{144}\)

∴ \(\text{AM} = 12\) cm ... (i)

Now, OM \(\perp\) AB ... (Given)

∴ \(\text{AB} = 2\text{AM}\) ... (Theorem)

∴ \(\text{AB} = 2 \times 12 = 24\) cm ... (ii)

Now, the chords of a circle equidistant from the centre are congruent. ... (Theorem)

∴ PQ = AB

∴ PQ = 24 cm ... (iii)

The lengths of the chords are 24 cm each.

Proof:

Draw seg CM and seg CN.

 In \(\triangle\)CMP and \(\triangle\)CNP,

  seg CM = seg CN ... (Radii of the same circle)

  seg PM = seg PN ... (Given)

  seg CP = seg CP ... (Common side)

∴ \(\triangle\)CMP \(\cong\) \(\triangle\)CNP ... (SSS Test)

∴ \(\angle\)MPC = \(\angle\)NPC ... (c. a. c. t.)

∴ Ray PC is the bisector of \(\angle\)NPM. ... (Definition of Angle Bisector)