Length of a Chord

By thales theorem angle ACB is a right angle and because OH is parallel with CB then OH is perpendicular to AC. And because AC is a chord and OH is perpendicular to the chord we know that S is the midpoint of AC. Now we will calculate lenght AC by using pythagors theorem: $|AC|=\sqrt{10^2-1,25^2}=\frac{5\sqrt{63}}{4}$ The lenght of OS is half the lenght CB, because it is a midline of triangle ABC, thus the length of SH is $\frac{45}{8}$ and now we can calculate the length CH from right triangle CHS: $|CH|=\sqrt{\frac{25*63}{64}+\frac{45^2}{64}}=\frac{15}{2}$ and this is our answer.

Let $\overline {AB}$ and $\overline {CH}$ meet at $D$ , and $\angle {HCB}=α$ . Then applying sine rule to $\triangle {CBD}$ we have

$\dfrac{|\overline {DB}|}{\sin α}=\dfrac{|\overline {CD}|}{\sin 2α}=\dfrac{|\overline {CB}|}{\sin 3α}$ .

Applying sine rule to $\triangle {ODH}$ we have

$\dfrac{|\overline {DH}|}{\sin 2α}=\dfrac{|\overline {MH}|}{\sin 3α}=\dfrac{|\overline {MD}|}{\sin α}=\dfrac{5-|\overline {DB}|}{\sin α}$ .

From these we get $5\sin α=4\sin 3α\implies \cos α=\dfrac{3}{4}\implies |\overline {DH}|=6, |\overline {CD}|=1.5, |\overline {CH}=6+1.5=\boxed {7.5}$ .

Let $AB$ and $CH$ intersect at $D$ and $OD = d$ ; then $DB = 5-d$ . We note that $\triangle BCD$ and $\triangle OHD$ are similar. Therefore, $\dfrac {OD}{DB} = \dfrac d{5-d} = \dfrac {OH}{CB} = \dfrac 5{1.25} = 4 \implies d = 20-4d \implies d=4$ . Similarly, let $CD=a$ ; then $HD = 4a$ . By intersecting chord theorem we have:

$\begin{aligned} CD\cdot HD & = DB \cdot AD \\ a \cdot 4a & = (5-4)(5+4) \\ 4a^2 & = 9 \\ \implies a & = \frac 32 \\ CH & = 5a = \boxed{7.5} \end{aligned}$