Please solve q 38 ?

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Answer 1 · 2018-07-01T15:24:29

#rarrBE=EC# As E is the mid point of BC.

In rt #Delta ABE#,

#rarrAB^2=AE^2+BE^2#

In rt #DeltaADE#,

#rarrAD^2=AE^2+ED^2#

#LHS=AB^2-AD^2#

#=AE^2+BE^2-(AE^2+ED^2)#

#=cancel(AE^2)+BE^2cancel(-AE^2)-ED^2#

#=BE^2-ED^2#

#=(BE+ED)(BE-ED)#

#=(CE+ED)*BD=CD*BD=RHS#