Ab is a diameter of a …

Clearly,{tex}\angle M P O = 90 ^ { \circ }{/tex}
Since {tex}B K \perp H M , A H \perp H M \text { and } O P \perp H M{/tex}. Therefore, AH||OP||BK .
Let AH = x, BK = y and OP = r. Further, let  MB = z.
In {tex}\Delta M K B{/tex} and {tex}\Delta M H A{/tex},we have
{tex}\angle M K B = \angle M H A{/tex}= 90°
and {tex}\angle B M K = \angle A M H{/tex} [Common]
So, by AA criterion of similarity, we have
{tex}\Delta M K B \sim \Delta M H A{/tex}
 {tex}\Rightarrow \quad \frac { B K } { A H } = \frac { M B } { M A }{/tex}
{tex}\Rightarrow \frac { y } { x } = \frac { z } { 2 r + z }{/tex}
{tex}\Rightarrow 2 r y + y z = z x{/tex}
{tex}\Rightarrow z = \frac { 2 r y } { x - y }{/tex}...(i)
In {tex}\Delta M K B{/tex} and {tex}\Delta M P O{/tex}, We have 
{tex}\angle M K B = \angle M P O = 90 ^ { \circ }{/tex}
{tex}\angle B M K = \angle O M P{/tex} [Common]
So, by AA criterion of similarly, we obtain
{tex}\Delta M K B \sim \Delta M P O{/tex}
{tex}\Rightarrow \quad \frac { B K } { O P } = \frac { B M } { O M }{/tex}
{tex}\Rightarrow \frac { y } { r } = \frac { z } { r + z }{/tex}
{tex}\Rightarrow r y + y z = r z{/tex}
{tex}\Rightarrow \quad z = \frac { ry } { r - y }{/tex}...(ii)
From (i) and (ii), we get
{tex}\frac { r y } { r - y } = \frac { 2 r y } { x - y }{/tex}
{tex}\Rightarrow 2 r - 2 y = x - y{/tex}
{tex}\Rightarrow 2 r = x + y{/tex}
Thus, AB = AH +BK.