ABC is a right angled triangle with AB=4, BC=3, AC=5. P is midpoint of AB . Q is a point on AC such that $\angle \mathrm{PQA}=90. \mathrm{BA}=\frac{4}{\mathrm{a}}\sqrt{13}$ then a =_____

$$\begin{gathered} \mathrm{PC}=\sqrt{3^2+2^2}=\sqrt{13} \\ ∆ \mathrm{AQP}, ∆\mathrm{ABC} are similar \end{gathered}$$ 

$$\begin{gathered} \frac{\mathrm{PQ}}{\mathrm{BC}}=\frac{\mathrm{PA}}{\mathrm{AC}}\Rightarrow \mathrm{PQ}=\frac{3\times 2}{5}=\frac{6}{5} \\ in ∆\mathrm{PQC}\Rightarrow {\mathrm{CP}}^2={\mathrm{PQ}}^2+{\mathrm{CQ}}^2 \\ \Rightarrow CQ=\sqrt{13-\frac{36}{25}}=\frac{17}{5} \\ \mathrm{BPQC} \mathrm{is} \mathrm{cyclic} \mathrm{quadrilateral} \end{gathered}$$

By Ptolemy’s theorem 
$$\begin{gathered} \mathrm{BQ}\times \mathrm{CP}=\mathrm{BP}\times \mathrm{CQ}+\mathrm{PQ}\times \mathrm{BC} \\ \mathrm{BQ}=\frac{2\left({\displaystyle \frac{17}{5}}\right)+\left({\displaystyle \frac{6}{5}}\right)3}{\sqrt{13}}=\frac{52}{5\sqrt{13}}=\frac{4}{5}\sqrt{13} \end{gathered}$$