If the length of the string is fixed (i.e. if the string is inextensible), then we have :
($l$ is the length of the string):
$$l=(s_1-s_3)+(s_1-s_5)+\pi R=2s_1-s_3-s_5+\pi R$$ $$l=(s_2-s_4)+(s_2-s_6)+\pi R=2s_2-s_4-s_6+\pi R$$ $$z=s_2-s_1\quad x=s_4-s_3\quad y=s_6-s_5$$ $$\Longrightarrow\; 2s_2-2s_1=(s_4-s_3)+(s_6-s_5)$$ $$\Longrightarrow\; 2z=x+y$$
If the length of the string is fixed (i.e. if the string is inextensible), then we have ($l$ is the length of the string):
$$l=(s_1-s_3)+(s_1-s_5)+\pi R=2s_1-s_3-s_5+\pi R$$ $$l=(s_2-s_4)+(s_2-s_6)+\pi R=2s_2-s_4-s_6+\pi R$$ $$z=s_2-s_1\quad x=s_4-s_3\quad y=s_6-s_5$$ $$\Longrightarrow\; 2s_2-2s_1=(s_4-s_3)+(s_6-s_5)$$ $$\Longrightarrow\; 2z=x+y$$
If the length of the string is fixed (i.e. if the string is inextensible), then we have:
($l$ is the length of the string)
$$l=(s_1-s_3)+(s_1-s_5)+\pi R=2s_1-s_3-s_5+\pi R$$ $$l=(s_2-s_4)+(s_2-s_6)+\pi R=2s_2-s_4-s_6+\pi R$$ $$z=s_2-s_1\quad x=s_4-s_3\quad y=s_6-s_5$$ $$\Longrightarrow\; 2s_2-2s_1=(s_4-s_3)+(s_6-s_5)$$ $$\Longrightarrow\; 2z=x+y$$
I think there is a typographical mistake in your question. I guess you want a proof for $\dot z=\large{\frac{\dot x+\dot y}2}$ or $v_O=\large{\frac{v_A+v_B}2}$.
If the length of the string is fixed (i.e. if the string is inextensible), then we have ($l$ is the length of the string):
$$l=(z-x)+(z-y)+2R=2z-(x+y)+2R$$
$$l=(s_1-s_3)+(s_1-s_5)+\pi R=2s_1-s_3-s_5+\pi R$$ $$\Longrightarrow\; 2z=x+y+\textrm{constant}$$$$l=(s_2-s_4)+(s_2-s_6)+\pi R=2s_2-s_4-s_6+\pi R$$ $$\Longrightarrow\; \frac d{dt}(2z)=\frac d{dt}\left(x+y+\textrm{constant}\right)$$$$z=s_2-s_1\quad x=s_4-s_3\quad y=s_6-s_5$$ $$\Longrightarrow\; 2\dot z=\dot x+\dot y\quad \text{or}\quad 2v_O=v_A+v_B$$$$\Longrightarrow\; 2s_2-2s_1=(s_4-s_3)+(s_6-s_5)$$ $$\Longrightarrow\; 2z=x+y$$
I think there is a typographical mistake in your question. I guess you want a proof for $\dot z=\large{\frac{\dot x+\dot y}2}$ or $v_O=\large{\frac{v_A+v_B}2}$.
If the length of the string is fixed (i.e. if the string is inextensible), then we have ($l$ is the length of the string): $$l=(z-x)+(z-y)+2R=2z-(x+y)+2R$$ $$\Longrightarrow\; 2z=x+y+\textrm{constant}$$ $$\Longrightarrow\; \frac d{dt}(2z)=\frac d{dt}\left(x+y+\textrm{constant}\right)$$ $$\Longrightarrow\; 2\dot z=\dot x+\dot y\quad \text{or}\quad 2v_O=v_A+v_B$$
If the length of the string is fixed (i.e. if the string is inextensible), then we have ($l$ is the length of the string):
$$l=(s_1-s_3)+(s_1-s_5)+\pi R=2s_1-s_3-s_5+\pi R$$ $$l=(s_2-s_4)+(s_2-s_6)+\pi R=2s_2-s_4-s_6+\pi R$$ $$z=s_2-s_1\quad x=s_4-s_3\quad y=s_6-s_5$$ $$\Longrightarrow\; 2s_2-2s_1=(s_4-s_3)+(s_6-s_5)$$ $$\Longrightarrow\; 2z=x+y$$
I think your question has there is a typographical mistake in your question. I guess you want a proof for $\dot z=\large{\frac{\dot x+\dot y}2}$ or $v_O=\large{\frac{v_A+v_B}2}$.
If the length of the string is fixed (i.e. if the string is inextensible), then we have ($l$ is the length of the string): $$l=(z-x)+(z-y)+2R=2z-(x+y)+2R$$ $$\Longrightarrow\; 2z=x+y+\textrm{constant}$$ $$\Longrightarrow\; \frac d{dt}(2z)=\frac d{dt}\left(x+y+\textrm{constant}\right)$$ $$\Longrightarrow\; 2\dot z=\dot x+\dot y\quad \text{or}\quad 2v_O=v_A+v_B$$
I think your question has a typographical mistake. I guess you want a proof for $\dot z=\large{\frac{\dot x+\dot y}2}$ or $v_O=\large{\frac{v_A+v_B}2}$.
If the length of the string is fixed (i.e. if the string is inextensible), then we have ($l$ is the length of the string): $$l=(z-x)+(z-y)+2R=2z-(x+y)+2R$$ $$\Longrightarrow\; 2z=x+y+\textrm{constant}$$ $$\Longrightarrow\; \frac d{dt}(2z)=\frac d{dt}\left(x+y+\textrm{constant}\right)$$ $$\Longrightarrow\; 2\dot z=\dot x+\dot y\quad \text{or}\quad 2v_O=v_A+v_B$$
I think there is a typographical mistake in your question. I guess you want a proof for $\dot z=\large{\frac{\dot x+\dot y}2}$ or $v_O=\large{\frac{v_A+v_B}2}$.
If the length of the string is fixed (i.e. if the string is inextensible), then we have ($l$ is the length of the string): $$l=(z-x)+(z-y)+2R=2z-(x+y)+2R$$ $$\Longrightarrow\; 2z=x+y+\textrm{constant}$$ $$\Longrightarrow\; \frac d{dt}(2z)=\frac d{dt}\left(x+y+\textrm{constant}\right)$$ $$\Longrightarrow\; 2\dot z=\dot x+\dot y\quad \text{or}\quad 2v_O=v_A+v_B$$
