In the given Fig. DEFG is a square and ∠BAC = 90°. Prove that
(i) ∆AGF ~ ∆DBG.
(ii) ∆AGF ~ ∆EFC.
(iii) ∆DBG ~ ∆AEFC
(iv) DE2 = BD × EC.
In ∆ADE and ∆ABC
∠ADE = ∠ABC
[corresponding angles]
and    ∠A = ∠A    [common]
Therefore, by using AA similar condition
        Â
     Â
              ...(i)
[taking reciprocals of both sides]
It is given that,
       Â
     Â
     Â
                  [Adding '1' both side]
      Â
                  ...(ii)
Comparing (i) and (ii), we get