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.

(i) In ∆AGF and ∆DBG
∠A = ∠BDG = 90°
and    ∠AGF = ∠DBG
[corresponding angles]
Therefore, by using AA similar condition
∆AGF ~ ∆DBG.
(ii) In ∆AGF and ∆EFC
∠AFG = ∠ECF
[corresponding angles]
and    ∠FAG = ∠CEF = 90°
Therefore, by using A.A similar condition
∆AGF ~ ∆EFC
From (i), we have
∆AGF ~ ∆DBG
From (ii), we have
              
          
           
             [corresponding sides of two similar triangles are proportional]