A square DEFG is inscribed in a rt.triangle ABC rt.angled at C in such a manner that D & E lie on AB and F.G lie on BC and CA. Prove that DE2=AD.EB

Dear student

In CGF and DAG  , we haveGCF=ADG    [each 90°]and CGF=DAG     [Corresponding angles]CGF~DAG    [By AA]In CGF and EFB, we haveFCG=BEF    [Each 90]CFG=EBF    [Corresponding angle]CGF~EFB    [By AA]Since CGF~DAG and  CGF~EFBDAG~EFBADEF=DGEBADDE=DEEB      DEFG is a square]DE2=AD×EB
Regards

  • 0
What are you looking for?