In the given figure , PN and UL are respectively the bisectors of angle P and angle U of a quadrilateral JUMP meet MU and JP produced in N and L . Prove that : Angle JPM+ angle JUM = 2 ( angle N + angle L ) solution​

Dear Student,

From triangle JUL,
$\angle$J + $\angle$JUL + $\angle$L = 180 ............ (1) (Sum of angles of a triangle is equal to 180)
But, $\angle$JUL = $\frac{\angle \mathrm{JUM}}{2}$    ( Since LU is angle bisector)
therefore, (1) implies
$\angle$J + $\frac{\angle \mathrm{JUM}}{2}$ + $\angle$L = 180  ............... (2)
Now, From triangle PNM,

$\angle$N + $\angle$NPM + $\angle$M = 180  .......... (3) ​(Sum of angles of a triangle is equal to 180)

But, ​ $\angle$NPM = $\frac{\angle \mathrm{JPM}}{2}$      ​( Since PN is angle bisector)
(3) implies
$\angle$N + ​$\frac{\angle \mathrm{JPM}}{2}$  + ​$\angle$M  = 180 ................ (4)
(2) + (4) gives,

$\angle$J + $\frac{\angle \mathrm{JUM}}{2}$ + $\angle$L  + ​$\angle$N + ​$\frac{\angle \mathrm{JPM}}{2}$  + ​$\angle$M = 360  ............. (5)
 Now, 
$\angle$J + $\angle$M + $\angle$P + $\angle$U = 360       (Sum of angles of quadrilateral is 360)
(5) implies
$\frac{\angle \mathrm{JUM}}{2}$+$\frac{\angle \mathrm{JPM}}{2}$ + ​$\angle$L  + ​$\angle$N + 360 - ​$\angle$P - $\angle$U = 360 .............. (6)
but $\angle$P = $\frac{\angle \mathrm{JPM}}{2}$ and $\angle$U = $\frac{\angle \mathrm{JUM}}{2}$
(6) implies

$\angle$JPM + $\angle$JUM = 2 ($\angle$N + $\angle$L) 
Hence proved

Regards