Q.

ΔABC is right-angled at A and DEFG is a square as shown in the figure. Then:

(i) ΔCBG ~ ΔEFC (ii) BD × EC = DE²

 (iii) ΔAGF ~ ΔDBG (iv) ΔAGF ~ ΔEFC

Choose the correct option from the following:

1 (iii) and (iv) are correct.  
2 (i), (ii) and (iii) are correct.  
3 (ii), (iii) and (iv) are correct.  
4 (i) and (ii) are correct.  

Ans.

(3)

In ΔAGF and ΔDBG:

∠GAF = ∠BDG = 90° and

∠AGF = ∠DBG (Corresponding angles)

∴ ΔAGF ~ ΔDBG (By AA similarity) …(A)∴ (iii) is correct.

Similarly, in ΔAGF and ΔEFC: 

∠GAF = ∠CEF = 90° and

∠AFG = ∠FCE (Corresponding angles)

∴ ΔAGF ~ ΔEFC (By AA similarity) …(B)∴ (iv) is correct.

From (A) and (B):ΔDBG ~ ΔEFC

DBEF=DGEC

But EF = DG = DE (side of square)

DBDE=DEEC⇒DE2=DB×EC∴ (ii) is correct.