How show that $AB>BR$ in rhombus? Let $A,B$ be points on the side $QR$ and $C,D$ be points on the side $RS$ of the rhombus $PQRS$, such that $B$ is closer to $R$ than $A$ is and that $C$ is closer to $R$ than $D$ is. Suppose that the segments $PA$, $PB$, $PC$, $PD$ divide the angle $\angle SPQ$ into five equal angles.
How show that $AB>BR$?

connect $SR$, let $\angle BSR=x \implies \angle ASB=2x, \angle BRS=5x<\dfrac{\pi}{2} $
let $BR=p,AB=q,SR=a,SB=b$, apply sin law:
$\dfrac{p}{\sin{x}}=\dfrac{a}{\sin{6x}},\dfrac{q}{\sin{2x}}=\dfrac{b}{\sin{8x}},\dfrac{a}{\sin{6x}}=\dfrac{b}{\sin{5x}} \implies \dfrac{q}{p}=\dfrac{\sin{2x}\sin{5x}}{\sin{x}\sin{8x}}=f(x)$
$f(x)=\dfrac{2\sin{5x}\cos{x}}{2\sin{4x}\cos{4x}}=\dfrac{\sin{6x}+\sin{4x}}{2\sin{4x}\cos{4x}}=\dfrac{\sin{4x}\cos{2x}+\cos{4x}\sin{2x}}{2\sin{4x}\cos{4x}}+\dfrac{1}{2\cos{4x}}=\dfrac{\cos{2x}}{2\cos{4x}}+\dfrac{\sin{2x}}{2\sin{4x}}+\dfrac{1}{2\cos{4x}}=\dfrac{\cos{2x}}{2\cos{4x}}+\dfrac{1}{4\cos{2x}}+\dfrac{1}{2\cos{4x}}=\dfrac{2\cos^2{2x}+1}{4\cos{4x}\cos{2x}}+\dfrac{1}{2\cos{4x}}=\dfrac{\cos{4x}+2}{4\cos{4x}\cos{2x}}+\dfrac{1}{2\cos{4x}}=\dfrac{1}{4\cos{2x}}+\dfrac{1}{2\cos{4x}\cos{2x}}+\dfrac{1}{2\cos{4x}}$
note $2x<4x <5x <\dfrac{\pi}{2}, \dfrac{1}{\cos{2x}}$ and $\dfrac{1}{\cos{4x}}$ are mono increasing funtion and positive $\implies f(x)$ is mono increasing function $ \implies f_{min}=f(0)=\dfrac{5}{4} \implies \dfrac{q}{p}>\dfrac{5}{4}>1 $
