In the given figure, triangle LMN is an isosceles triangle with LM=LN.LP bisects angle NLQ.Prove that LP parallel to MN. Share with your friends Share 28 Manbar Singh answered this In ∆LMN, LM = LN Given⇒∠LNM = ∠LMN Angles opposite to equal sides are equal ....1Since, LP bisects ∠QLN, then∠QLP = ∠PLN = 12∠QLN⇒∠QLN = 2∠PLN ....2Now, in ∆LMN,ext. ∠QLN = ∠LNM + ∠LMN Exterior angle theorem⇒2∠PLN = ∠LNM +∠LNM Using 1 and 2⇒2∠PLN = 2∠LNM⇒∠PLN = ∠LNMBut ∠PLN and ∠LNM are alternate interior angles made by the transversal with lines LP and MN and are equal.So, LP∥MN 32 View Full Answer Ujjwal Gupta answered this Triangle LMN is an isosceles triangle with LM=LN So, base angles M=N=x Let the angle L be y So, y= 180-2x (1) If we extend the lines LP,MN and QM then QM is transversal of two horizontal lines LP and MN So angle M= angle QLP= x (corresponding angles) Since LP is bisector of angle NLQ angle QLP= angle PLN= x (2) If the sum of angles PLM and LMN is 180 then LP parallel to MN (via Euclid's Postulate 5) So, PLM+LMN= (PLN+NLM)+ LMN = (x+180-2x) + x = x+180-2x+x =2x-2x+180 = 180 Since angle PLM + angle LMN= 180 LP parallel to MN -2
