C-03-Q037mediumsingle_mcqIn △ABC\triangle ABC△ABC, ∠C=90°\angle C = 90°∠C=90° and DDD is the midpoint of BCBCBC. Then AB2=AB^2 =AB2=aAD2+BD2AD^2 + BD^2AD2+BD2bAD2+2BD2AD^2 + 2 BD^2AD2+2BD2cAD2+3BD2AD^2 + 3 BD^2AD2+3BD2d2(AD2+BD2)2(AD^2 + BD^2)2(AD2+BD2)