In the given figure, triangle LMN is an isosceles triangle with LM=LN.LP bisects angle NLQ.Prove that LP parallel to MN.

In LMN,    LM = LN   GivenLNM = LMN   Angles opposite to equal sides are equal    ....1Since, LP bisects QLN, thenQLP = PLN = 12QLNQLN = 2PLN    ....2Now, in LMN,ext. QLN = LNM + LMN   Exterior angle theorem2PLN = LNM +LNM    Using 1 and 22PLN = 2LNMPLN = LNMBut PLN and LNM are alternate interior angles made by the transversal with lines LP and MN and are equal.So, LPMN

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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)
​                          = (x+180-2x) + x
                          = x+180-2x+x
                          = 180
Since angle PLM + angle LMN= 180
LP parallel to MN

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