CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ≅ ΔFEG, show that: (i) CD/GH = AC/FG (ii) ΔDCB ≅ ΔHGE (iii) ΔDCA ≅ ΔHGF

In ΔABC and ΔFEG, ΔABC ≅ FEG ⇒ ∠ACB = ∠EGF (Corresponding angles of similar triangles)
Since, DC and GH are bisectors of ∠ACB and ∠EGH respectively.
⇒ ∠ACB = 2∠ACD = 2∠BCD
And ∠EGF = 2∠FGH = 2∠HGE
⇒ ∠ACD = ∠FGH and ∠DCB = ∠HGE.. (1)
Also ∠A = ∠F and ∠B = ∠E.. (2)
In ΔACD and ΔFGH, ∠A = ∠F (From 2) ∠ACD = ∠FGH (From 1)
⇒ By AA criterion of similarity ΔACD ≅ ΔFGH
ΔDCA ≅ ΔHGF [(i) and (iii) proved]
⇒ CD/GH = AC/FG (Corresponding Sides of Similar Triangles)
In ΔDCB and ΔHGE, ∠B = ∠E (From 2) ∠DCB = ∠HGE (From 1)
⇒ By AA criterion of similarity ΔDCB ≅ ΔHGE [(ii) proved]