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 $DeltaADE$ ,

$rarrAD^2=AE^2+ED^2$

$LHS=AB^2-AD^2$

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

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

$=BE^2-ED^2$

$=(BE+ED)(BE-ED)$

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