<s>[INST] What's the measure of the segment $PQ$ in the triangle below? For reference:
NIn figure CQ is bisector, and
PE = EC, AC = b and BC = a. Calculate PQ.(Answer: $\frac{ab}{a+b}$)

My progress:
$(Bissector ~Th.):\triangle ABC:\\ \frac{BQ}{AQ}=\frac{a}{b} \\ 
By~ proportionality: \frac{BQ}{BQ+AQ}=\frac{a}{a+b}=\frac{BQ}{AB}$
$(Bissector~ Th.):\triangle CEP:\\  \frac{PQ}{EQ}=\frac{CP}{CE}\\ 
By~ proportionality: \frac{PQ}{PQ+EQ}=\frac{CP}{CE+CP}\implies \frac{PQ}{PE=EC} = \frac{CP}{CE+CP}$
$(Menelaus~ Th.): CEP-AB:\\: a \cdot PQ \cdot AE =BP \cdot EQ \cdot b\implies PQ = \frac{b}{a} \cdot.\frac{BP \cdot EQ}{AE}(I) \\ (Menelaus - Th.): \triangle ABC - PE:\\CE \cdot AQ \cdot BP = AE \cdot BQ \cdot CP \implies \frac{BP}{AE} = \frac{a}{b}\cdot\frac{CP}{CE}(II)\\
(II)in (I): PQ = \frac{CP}{CE} \cdot EQ \implies PQ = \frac{(a+PB) \cdot (b-PQ)}{(b-EA)}$
 [/INST] 
Draw segment $QR = QP$ then,
$\angle PRQ = \angle PCA \implies AC \parallel QR$. Also, $CQ$ is angle bisector of $\angle ACB$.
So, $ \displaystyle \frac{QR}{AC} = \frac{BQ}{BA} = \frac{BC}{BC+AC} = \frac{a}{a+b}$
$ \displaystyle \implies PQ = QR = \frac{ab}{a+b}$
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