An equation involving ratios in a triangle. In triangle $ABC$, if the incenter is $I$ and $AI$ meets $BC$ at $D$, show that
$$\frac{AD}{ID}=\frac{AB+BC+CA}{BC}$$
I tried using similar triangles and got nowhere, couldn't find any use for the fact that the numerator of the RHS is the perimeter either .

Responding to @Blue's comment in attempt to complete it, if $r$ is the inradius, 
$$RHS = \frac{\frac12r(AB+BC+CA)}{\frac12r\cdot BC}=\frac{|\triangle IAB|+|\triangle IBC|+|\triangle ICA|}{|\triangle IBC|}=\frac{|\triangle ABC|}{|\triangle IBC|}=\frac{AP}{IQ}$$
where  $P,Q$ are the feet of perpendiculars from $A, I$ on $BC$.
But $AP \parallel IQ \implies \triangle APD \sim \triangle IQD \implies LHS=\dfrac{AD}{ID}=\dfrac{AP}{IQ}=RHS$.
