ABCD is a parallelogram and AL and CM are perpendicualrs from vertices A and C on diagonal BD, as shown in the adjoining figure. Show that (i) `triangle ALB ~= triangle CMD` and (ii) AL = CM.

Explanation

In `triangle ALB and CMD, ` we have
`AB =CD " " ` (opp. sides of a ||gm)
`angle ALB = angle CMD =90^(@) " " [ because AL bot BD, CM bot BD]`
`angle ABL = angle CDM " " ("alt. interior " angle)`
`therefore triangle ALB ~= triangle CMD " " ` (AAS - criterion).
And so, `AL = CM " " ` (c.p.c.t.).