A transversal cuts the sides OL,OM and diagonal ON of a parallelogram at A,B and C respectively.
Prove that `(OL)/(OA)+(OM)/(OB)=(ON)/(OC)`.

Options

A. Statement - 1 is True, Statement - 2 is True , Statement - 2 is a correct explanation for Statement - 1.
B. Statement -1 is True, Statement - 2 is True, Statement -2 is not a correct explanation for Statement - 1.
C. Statement - 1 is True, Statement - 2 is False.
D. Statement - 1 is False, Statement - 2 is True.

Answer

A

Explanation

statement -2 is true
Let `vec(OL) = x vec(OA), vec(OM) = y O vec(B) and vec(ON) = z vec(OC).` Then,
`|vec(OL)| = x|vec(OA)| , |vec(OM)|=y|vec(OB)| and |vec(ON)| = z | vec(OC)|`
`rArr x= (OL)/(OA), y=(OM)/(OB), z=(ON)/(OC)`

Talking O as the origin, let the position vectors of A, B, C be `vec a , vec b and vec c ` respectively.
In `Delta OLN,` we have
`vec(OL) +vec(LN)= vec(ON)`
`rArr vec(ON) = vec(OL)+vec(OM)`
`rArr z vec(OC) = x vec(OA) + y vec(OB)`
`rArr x vec(OA) +y vec(OB) -z vec(OC)=vec 0 rArr x vec(a) + y vec (b) + (-z) vec c = vec 0`
But, points A, B, C are collinear.
`therefore x vec a + y vec b + (-z) vec c = vec 0`
`rArr x+y-z =0 rArr x+y=z rArr (OL)/(OA) + (OM)/(OB) =(ON)/(OC)`
So, statement -1 is also true and statement -2 is a correct explanation for statement -1.