1consider the triangle AEF
√AEF=90-√BEF
=90-√FAD(by same angle in same segment rule)
=90-(90-B)=B
applying sine rule
x/sinA=bcosA/sinB
where x is a side of the interior triangle
x=bcosAsinA/sinB
=(b/sinB)cosAsinA
=2RcosAsinA
similary we get the semiperimeter as
R/2(sin2A +sin2B +sin2C)
further the inradius of the outer triangle=
4RsinA/2sinB/2sinC/2
equating both these
R/2(sin2A +sin2B +sin2C)=4RsinA/2sinB/2sinC/2
further sin2A+sin2B+sin2C=4sinAsinBsinC=
4*8sinA/2cosA/2sinB/2cosB/2sinC/2cosC/2
so we get cosA/2cosB/2cosC/2 as =1/4