Question

In a ΔABC, if L and M are the points on AB and AC respectively, such that LM ll BC. prove that ar(LOB)=ar(MOC)​

Answer 1

Step-by-step explanation:

angle LOB = angle MOC ( vertically opposite angle)

LB = MC

since midpoint divides into equal halves

angle LBO = angle MCO

since BM and CL bisects ABC and ACB into two equal halves

therefore LOB = MOC

Answer 2

Given, LM || BC

To prove, ar(LOB) = ar(MOC)

Proof

LM || BC and BC is the common base.

ar(LCB) = ar(MBC) (Triangles on the same base and between the same parallels are equal in area.)

Subtracting ar(BOC) on both sides.

ar(LCB)-ar(BOC) = ar(MBC) -ar(BOC)

ar(LOB) = ar(MOC)