Let $E \equiv PB \cap CR.$ $\triangle EBC$ is obviously equilateral. Since $\widehat{PQR}=120^{\circ}=90^{\circ}+\tfrac{1}{2}\widehat{PER}$ and $EQ$ bisects $\widehat{PER}$ internally, we deduce that $Q$ is the incenter of $\triangle EPR.$ Hence, if $M \equiv PQ \cap ER$ and $N \equiv RQ \cap EP,$ then by angle bisector theorem, using $QB \parallel ER,$ we get $\tfrac{BC}{AB}=\tfrac{BQ}{PB}=\tfrac{EM}{PE}=\tfrac{RM}{PR}.$ Similarly $\tfrac{BC}{CD}=\tfrac{PN}{PR},$ hence it is enough to show that $PN+RM=PR.$ Clearly $Q$ is midpoint of the arc $MN$ of $\odot(EMN),$ i.e. $\triangle QMN$ is Q-isosceles with $\widehat{MQN}=120^{\circ}.$ Thus, reflection $X$ of $M$ on $RQ$ forms equilateral triangle $\triangle XMN$ with center $Q$ $\Longrightarrow$ $X$ is also reflection of $N$ on $PQ.$ Hence, $RM=RX$ and $PN=PX$ $\Longrightarrow$ $PR=PX+RX=PN+RM,$ as desired.

Nice solution, Luis. Here is a trigonometric proof. Set $x=AB$, $y=BC$, $z=CA$. By the law of cosines we have \begin{align*} PQ^2&=x^2+y^2-2xy\cos 60^\circ=x^2+y^2-xy\\ QR^2&=y^2+z^2-2yz\cos 60^\circ=y^2+z^2-yz. \end{align*} Moreover, a quick application of the Pythagorean theorem allows us to write $PR$ in terms of $x$, $y$ and $z$: \begin{align*}PR^2&=\left(y+\frac{x+z}{2}\right)^2+\left(\frac{\sqrt{3}}{2}(z-x)\right)^2\\ &=x^2+y^2+z^2+xy+yz-xz.\end{align*} On the other hand, the condition $\angle PQR=120^\circ$ together with the law of cosines yields the following constraint on $PQ$, $QR$ and $PR$: \[PR^2=PQ^2+QR^2-2PQ\cdot QR \cos 120^\circ=PQ^2+QR^2+PQ\cdot QR.\] Substituting the first three equations we get \[ x^2+y^2+xy+yz-zx=x^2+y^2-xy+y^2+z^2-yz+\sqrt{(x^2+y^2-xy)(y^2+z^2-yz)}\] which is equivalent to \[2xy+2yz-xz=\sqrt{(x^2+y^2-xy)(y^2+z^2-yz)}.\] Squaring both sides of the last equation and rearranging we arrive at \[3x^2y^2+3y^2z^2+9y^2zx-3x^2yz-3xy^3-3xyz^2-3y^3z=0.\] The left-hand side factorizes as $3y(x-y+z)(xy+yz-zx)$, which means that either $y=0$, $x-y+z=0$ or $xy+yz-zx=0$. The first case is discarded since $y>0$, and the second one is impossible given the condition $x,z>y$. Hence $xy+yz-zx=0$, i.e. $1/x+1/z=1/y$, which is just what we want.

1. The circuncircle of $QCR$ meets $AB$ at $E$ 2. Note that $QRE$ is equilateral, then by Ptolemy's theorem $CR=QC+CE$ or $CD=BC+CE$ 3. Since $ \widehat{PQR}=120^{\circ}$ we have $\triangle QMN \sim \triangle CRE$ $\Longrightarrow$ $CE=BC*CD/AB$ 4. Then $CD=BC+BC*CD/AB \Longrightarrow \frac{1}{AB}+\frac{1}{CD}=\frac{1}{BC}$

\[\begin{gathered} Let\mathop {}\limits^{} E = PQ \cap AD\mathop {}\limits^{} and\mathop {}\limits^{} Z = RQ \cap AD. \hfill \\ we\mathop {}\limits^{} have\mathop {}\limits^{} : \hfill \\ Z\hat QE = {120^o}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} \theta + \varphi = {60^o} \hfill \\ {{\hat Z}_1} + \theta = {60^o}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} {{\hat Z}_1} = \varphi \hfill \\ {{\hat Z}_1} = {{\hat {\rm P}}_1}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} P,Z,B,Q\mathop {}\limits^{} is\mathop {}\limits^{} concyclic\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} {{\hat Z}_2} = {60^o}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} ZP||ER\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} \hfill \\ EZPR\mathop {}\limits^{} is\mathop {}\limits^{} trapezium. \hfill \\ Let\mathop {}\limits^{} MN||ZP\mathop {}\limits^{} then\mathop {}\limits^{} tr.BZR \approx tr.CQN \approx tr.EDR\mathop {}\limits^{} and\mathop {}\limits^{} QM = QN\mathop {}\limits^{} \Rightarrow \hfill \\ \left. \begin{gathered} \frac{{BC}}{{CD}} = \frac{{ZQ}}{{ZR}} \hfill \\ \frac{{BC}}{{AB}} = \frac{{QR}}{{ZR}} \hfill \\ \end{gathered} \right\}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} \frac{{BC}}{{CD}} + \frac{{BC}}{{AB}} = 1\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} \frac{1}{{CD}} + \frac{1}{{AB}} = \frac{1}{{BC}} \hfill \\ \end{gathered} \]

Attachments:

Let $m\angle BQP = \alpha$ and $ m\angle CQR = \beta$. It's easy to see that $\alpha + \beta = 180^\circ$. By sine law on $ \triangle BPQ$ we have $\frac{BQ}{BP}=\frac{BC}{AB}=\frac{\sin{\angle BPQ}}{\sin{\angle BQP}}= \frac{\sin {(120 - \alpha)}}{\sin {\alpha}}$ And on $\triangle CQR$ we have $\frac{CQ}{CP}=\frac{BC}{CD}=\frac{\sin{\angle CRQ}}{\sin{\angle CQR}}= \frac{\sin {(120 - \beta)}}{\sin {\beta}}$. Thus \[\frac{BC}{AB} + \frac{BC}{CD} = \frac{\sin {(120 - \alpha)}}{\sin {\alpha}} + \frac{\sin {(120 - \beta)}}{\sin {\beta}} = \frac{\sin {(\alpha + 60)}}{\sin {\alpha}} + \frac{\sin {(\alpha - 60)}}{\sin {(180 - \alpha)}}\] \[= \frac{2\sin {\alpha} \cos60^\circ}{\sin {\alpha}} = 1\]. Hence the result.

Let $S=\overline{PB}\cap\overline{RC}, U=\overline{PQ}\cap\overline{RS}, V=\overline{QR}\cap\overline{PS}$. Hence $Q$ is the incenter of $\triangle PSR$ and $\angle PSR=60^{\circ}$. No we focus on a small part of the problem Claim. RU+PV=PR Proof. Reflect $U$ over $\overline{QR}$ to $U'$. Hence $U'$ lies on $PR$ since $Q$ is an incenter. Hence \begin{align*} \angle PUQ'&=180^{\circ}-\angle QU'R\\ &=180^{\circ}-\angle QUR\\ &=180^{\circ}-\left(60^{\circ}+\frac{\angle P}{2}\right)\\ &=120^{\circ}-\frac{\angle P}{2}\\ &=60^{\circ}+\frac{120^{\circ}-\angle P}{2}\\ &=60^{\circ}+\frac{\angle R}{2}\\ &=\angle PVQ \end{align*}so $U'$ is the reflection of $V$ over $\overline{PQ}$. $$RU+PV=RU'+PU'=PR$$Hence \begin{align*} \frac{1}{AB}+\frac{1}{CD}&=\frac{1}{BC}\left(\frac{BC}{AB}+\frac{BC}{CD}\right)\\ &=\frac{1}{BC}\left(\frac{QB}{PB}+\frac{QC}{CB}\right)\\ &=\frac{1}{BC}\left(\frac{US}{PS}+\frac{VS}{SR}\right)\\ &=\frac{1}{BC}\left(\frac{UR}{PR}+\frac{VP}{PR}\right)\\ &=\frac{1}{BC} \end{align*}