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 .
 A: First, let's prove that in arbitrary triangle $XYZ$ with bisector $XT$, $T$ lays on the segment $YZ$: $$\frac{TY}{TZ}=\frac{XY}{XZ}\hbox{.}$$
Compare the two areas of $\Delta XYT$ and $\Delta XZT$:
$$\frac{S_{XYT}}{S_{XZT}}=\frac{YT}{ZT}$$
because $\Delta XYT$ and $\Delta XZT$ share common height.
Then, $$\frac{S_{XYT}}{S_{XZT}}=\frac{\frac{1}{2}XY\cdot XT\cdot \sin\angle YXT}{\frac{1}{2}XZ\cdot XT\cdot \sin\angle ZXT}=\frac{XY}{XZ},$$
hence the demanded ratio equality (I just thought it's well-known).
The bisectors meet in the incenter;
$$\Delta ADB:~~ \frac{ID}{AI}=\frac{BD}{AB}  $$
$$\Delta ADC:~~ \frac{ID}{AI}=\frac{CD}{AC}  $$
If $\frac{x}{y}=\frac{z}{t}$ then $\frac{x}{y}=\frac{z}{t}=\frac{x+z}{y+t}$, hence
$$\frac{ID}{AI}=\frac{BD+CD}{AB+AC}=\frac{BC}{AB+AC}$$
$$\frac{AI}{ID}=\frac{AB+AC}{BC}$$
$$\frac{AI}{ID}+1=\frac{AB+AC}{BC}+1,$$
hence the desired ratio.
A: 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 $AP \parallel IQ \implies \triangle APD \sim \triangle IQD \implies LHS=\dfrac{AD}{ID}=\dfrac{AP}{IQ}=RHS$.
