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)
mark it as the brainliest
when an another answer is added to this solution.................
hope it helps u !!!!!!!!!!