Let $KM$ and $CO$ cut $AB$ at $P$ and $L^*.$ $\angle OLK=\angle OLK,$ i.e. $LO,AB$ bisect $\angle MLK$ $\Longleftrightarrow$ $L(M,K,O,P)=-1$ $\Longleftrightarrow$ $C(B,A,L^*,P)=-1$ $\Longleftrightarrow$ $L \equiv L^*$ $\Longleftrightarrow$ $CO$ is C-altitude of $\triangle ABC.$ By Ceva's theorem, we get $\frac{AL}{LB} \cdot \frac{BM}{MC}\cdot \frac{CK}{KA}=1 \Longleftrightarrow \frac{AL}{LB}=\frac{CL \cdot \cot A}{CL \cdot \cot B}=\frac{KA}{CK}=\frac{c}{a} \Longleftrightarrow$ $a(b^2+c^2-a^2)-c(a^2+c^2-b^2)=0 \Longleftrightarrow b= \sqrt{\frac{a^3+c^3+ca^2-ac^2}{a+c}}$

My ugly solution Let the line OC intersect AB in point P. As AM is a median, we have |AP| |PB| = |AK| |KC| (this obviously holds if |AB| = |AC| and the equality is preserved under uniform compression of the plane along BK). Applying the sine theorem to the triangles ABK and BCK we obtain $|AP|/|PB =|AK/KC|=|AB/BC|=5/4$ As |AP| + |PB| = |AB| = 15, we have |AP| = 25/3 and |PB| = 20/3 Thus $|AC|^2 -|BC|^2 = 25 = |AP|^2 -|BP|^2$ and $|AC|^2 -|AP|^2 = |BC|^2 |BP|^2$. Applying now the cosine theorem to the triangles APC and BPC we get cos\APC = cos\BPC, i.e., P = L. As above, we can use a compression of the plane to show that KP k BC and therefore \OPK = \OCB. As |BM| = |MC| and \BPC = 90 we have \OCB = \OPM. Combining these equalities, we get \OLK = \OPK = \OCB = \OPM = \OLM.

Who's Know P = L ?