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you can see the diagram given below.


Given: In a ΔABC

l is a straight line passing through the vertex A . BM ⊥ l and CN ⊥ l. L is the mid point of BC.

To prove: LM = LN

Construction: Draw OL ⊥ l


If a transversal make equal intercepts on three or more parallel lines, then any other transversal intersecting them will also make equal intercepts.

BM ⊥ l, CN ⊥ l and OL ⊥ l.

∴ BM || OL || CN

Now, BM | OL || CN and BC is the transversal making equal intercepts i.e., BL = LC.

∴ The transversal MN will also make equal intercepts.

⇒ OM = ON

In Δ LMO and Δ LNO,

OM = ON  

∠LOM = ∠LON  (OL is perpendicular to BC)

OL = OL    (Common line )

∴ ΔLMO ≅ ΔLNO  (By SAS congruence criterion)

∴ LM = LN ( By CPCT)


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Well, it is an easy problem.what u have to do is just complete the trapezium BMCN. now BM || CN. then u construct a perpendicular from L to the line and extend it such that it intersects BN at X, CM at Y and MN at K. now applying midpoint theorem to the triangles BCM and BCN ( using the fact that XY || BM || CN and L is the midpoint of BC, you can easily prove that K is the midpoint of MN. now in triangle LMN, LK is perpendicular to MN and bisects it which is possible only when the triangle is isoceles. therefore LM=LN.
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