2. As shown in the figure, in $\triangle ABC$, $\angle C=90^{\circ}$, points $P, Q$ are on the hypotenuse $AB$, satisfying the conditions $BP = BC = a$, $AQ = AC = b$, $AB = c$, and $b > a$. Draw $PM \perp BC$ at $M$, $QN \perp AC$ at $N$, and $PM$ and $QN$ intersect at $L$. Given that $\frac{S_{\triangle PQL}}{S_{\triangle ABC}}=\frac{4}{25}$, then $a: b: c=$
2. $3: 4: 5$ From similar triangles, we get $5(a+b)=7c$, squaring and rearranging gives $12a^2 - 25ab + 12b^2 = 0$, solving for $\frac{a}{b} = \frac{3}{4}$ or $\frac{4}{3}$ (discard), hence $a: b: c = 3: 4: 5$.