Question

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

Answer 1

PN is the bisector of ∠P.
∴, ∠JPN=∠NPM ------------------(1)
UL is the bisector of ∠U.
∴, ∠JUL=∠LUM -------------------(2)
∴, ∠JPM+∠JUM
=(∠JPN+∠NPM)+(∠JUL+∠LUM)
=2∠JPN+2∠LUM  [using (1) and (2)]
=2∠PNM+2∠JLU
[∵, from the given figure, JL||NM, PN traversal
∴, ∠JPN=∠PNM (alternate angles) and 
∵, JL||NM, UL traversal
∴, ∠JLU=∠LUM (alternate angles)]
=2(∠N+∠L) (Proved)

Answer 2

From triangle JUL, ∠J + ∠JUL + ∠L = 180 ............ (1) (Sum of angles of a triangle is equal to 180) But, ½∠JUL = ½∠JUM ( Since LU is angle bisector) therefore, (1) implies ∠J + ½∠JUM + ∠L = 180 ............... (2) Now, From triangle PNM, ∠N + ∠NPM + ∠M = 180 .......... (3) ​(Sum of angles of a triangle is equal to 180) But, ​ ∠NPM = ½∠JPM ​( Since PN is angle bisector) (3) implies ∠N + ​½∠JPM + ​∠M = 180 ................ (4) (2) + (4) gives, ∠J + ½∠JUM + ∠L + ​∠N + ​½∠JPM + ​∠M = 360 ............. (5) Now, ∠J + ∠M + ∠JPM + ∠JUM = 360 (Sum of angles of quadrilateral is 360)…… (6) From (5 ) and (6) ∠J + ½∠JUM + ∠L + ​∠N + ​½∠JPM + ​∠M = ∠J + ∠M + ∠JPM + ∠JUM (Angle M and J gets subtracted) ½∠JUM + ∠L + ​∠N + ​½∠JPM = ∠JPM + ∠JUM ∠L + ​∠N =∠JPM + ∠JUM - ½(∠JUM + ∠JPM) ∠L + ​∠N = ½(∠JUM + ∠JPM) 2( ∠L + ​∠N)= ∠JUM + ∠JPM Hence, proved.