Question

BM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. If
L is the mid-point of BC, prove that LM = LN.

Answer 1

To prove LM = LN
Draw LS perpendicular to line MN

∴ The lines BM, LS and CN being the same perpendiculars, on line MN are parallel to each
other.
According to intercept theorem,
If there are three or more parallel lines and the intercepts made by them on a transversal or
equal. Then the corresponding intercepts on any other transversal are also equal.
In the drawn figure, MB and LS and NC are three parallel lines and the two transversal line
are MN and BC
We have, BL= LC  (As L is the given midpoint of BC)
∴ using intercept theorem, we get

MS = SN               ....(i )

Now in Δ MLS and LSN

MS = SN using                ….(i)

${\displaystyle \angle }$LSM = ${\displaystyle \angle }$LSN = 90°LS ^ MN and SL = LS common

∴ Δ DMLS ≅  Δ LSN       (SAS congruency theorem)

∴ LM = LN   (CPCT )