Let [tex]x=m\widehat{EY}=m\widehat{YI}[/tex]. By the inscribed angle theorem, we have
[tex]m\angle EJI=\dfrac12m\angle EOI=\dfrac12m\widehat{EI}=\dfrac{m\widehat{EY}+m\widehat{YI}}2=x^\circ[/tex]
Then
[tex]m\angle KJE=(180-x)^\circ[/tex]
Also by the inscribed angle theorem, we have
[tex]m\angle ELY=\dfrac12m\angle EOY=\dfrac12m\widehat{EY}=\left(\dfrac x2\right)^\circ[/tex]
so that
[tex]m\angle KLX=\left(180-\dfrac x2\right)^\circ[/tex]
Angles EXY and LXJ form a vertical pair so they are congruent and both have measure [tex]80^\circ[/tex].
The sum of the interior angles to any quadrilateral is [tex]360^\circ[/tex], so for quadrilateral KLXJ we get
[tex]25^\circ+\left(180-\dfrac x2\right)^\circ+(180-x)^\circ+80^\circ=360^\circ\implies x^\circ=70^\circ[/tex]
So,
[tex]m\widehat{EI}=2x^\circ=\boxed{140^\circ}[/tex]
