(Q28) CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆FEG respectively. If ∆ABC ∼ ∆FEG then show that

i)

CD

GH

=

AC

FG

ii) ∆DCB ∼ ∆HGE

iii) ∆DCA ∼ ∆HGF

Solution :

Given :

∆ABC ∼ ∆FEG.

CD is the bisector of ∠C and GH is the bisector of ∠G.

R.T.P :

In ∆ACD and ∆FGH

∠A = ∠

[∵ Corresponding angles of ∆ABC and ∆FEG]

∠ACD = ∠FGH

∴ By A.A. similarity condition, ∆ACD ∼ ∆FGH

[∵ Ratio of the Corresponding angles is equal]

ii) ∆DCB ∼ ∆HGE

In ∆DCB and ∆HGE,

∠B = ∠

[∵ Corresponding angles of ∆ABC and ∆FEG]

∠DCB = ∠HGE

∴ ∆DCB ∼ ∆HGE . (by A.A. similarity condition)

iii) ∆DCA ∼ ∆HGF

In ∆DCA and ∆HGF

∠A = ∠