# BM and CN are perpendiculars to a line passing through the vertex A of triangle ABC.

Question:

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

Solution:

To prove LM = LN

Draw LS as perpendicular to line MN.

Therefore, 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 are equal, then the corresponding intercepts on any other transversal are also equal.

In the figure, MB, LS and NC are three parallel lines and the two transversal lines are MN and BC.

We have, BL = LC   [As L is the given mid-point of BC]

Using the intercept theorem, we get

MS = SN  ..... (i)

Now in ΔMLS and ΔLSN

MS = SN using equation ... (i)

∠LSM = ∠LSN = 90° [LS ⊥ MN]

And SL = LS is common.

∴  ΔMLS ≅ ΔLSN    [SAS Congruency Theorem]

∴  LM = LN           [CPCT]