Given $\triangle ABC\ :\ A(0,a),B(-b,0),C(c,0)$. Circumcenter $O(\frac{c-b}{2},\frac{a^{2}-bc}{2a})$. The tangent line in $B\ :\ y=\frac{a(b+c)}{bc-a^{2}}(x+b)$ intersects $y=a$ in the point $P(-\frac{a^{2}+b^{2}}{b+c},a)$. Point $Q(\frac{(a^{2}+b^{2})[c(c^{2}+bc+c^{2})-a^{2}b]}{N},\frac{a(a^{2}+c^{2})(b+c)^{2}}{N})$. with $N=a^{4}+a^{2}(3b^{2}+4bc+3c^{2})+b^{4}+2b^{3}c+3b^{2}c^{2}+2bc^{3}+c^{4}$. Point $R(\frac{-a^{4}b-a^{2}(2b^{3}+b^{2}c-c^{3})+c(2b^{4}+4b^{3}c+5b^{2}c^{2}+3bc^{3}+c^{4})}{N},-\frac{a(b+c)^{4}}{N})$. $BQCR$ is a parallelogram if the midpoint of $BC$ is equal to the midpoint of $QR$, if $x_{Q}+x_{R}=b+c$ and $y_{Q}+y_{R}=0$. This is only possible if $a^{2}=b^{2}+2bc$, if $a^{2}+c^{2}=(b+c)^{2}$, if $\sqrt{a^{2}+c^{2}}=b+c$, if $AC=BC$.

Synthetic solution: note that we always have $CQ \parallel BR$ (from an easy angle chasing). Thus $BQCR$ is parallelogram iff $CR \parallel BQ$, i.e. $\angle CRA=\angle BQR$. But we have \[ \angle CRA = \angle BQR \iff \angle ABC = \angle APB \iff \angle ABC=\angle CAB \iff AC=BC \]