In the adjoining figures, and
are perpendiculars to the diagonal
of a ||gm
Prove that
(i) (ii)
Here, ABCD is parallelogram.
Hence, AD || BC and AD = BC
(i) In ∆ALD and ∆CMB, we have,
AD = BC
∠ALD = ∠CMB (90o each)
∠ADL = ∠CBM (Alternate interior angle)
∴ ∆ALD ≅ ∆CMB
(ii) As ∆ALD ≅ ∆CMB …from 1
∴ AL = CM …by cpct