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


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