CD and GH are respectively the bisectors of Δ ACB and Δ EGF such that D and H lie on sides AB and FE of Δ ABC and Δ EFG respectively. If Δ ABC ~ Δ FEG, show that:

(i) $\dfrac{CD}{GH} = \dfrac{AC}{FG}$

(ii) Δ DCB ~ Δ HGE

(iii) Δ DCA ~ Δ HGF

(i) Given,

∆ ABC ~ ∆ FEG

We know that,

If two triangles are similar, then their corresponding angles are congruent.

⇒ ∠ACB = ∠FGE

⇒ $\dfrac{∠ACB}{2} = \dfrac{∠FGE}{2}$

⇒ ∠ACD = ∠FGH (CD and GH are bisectors of ∠C and ∠G respectively) …………..(1)

In ∆ ADC and ∆ FHG,

⇒ ∠DAC = ∠HFG [∵ ∆ABC ~ ∆FEG]

⇒ ∠ACD = ∠FGH [From equation (1)]

∴ ∆ ADC ~ ∆ FHG (By A.A. axiom)

We know that,

Ratio of corresponding sides in similar triangle are proportional.

∴ $\dfrac{CD}{GH} = \dfrac{AC}{FG}$.

Hence, proved that $\dfrac{CD}{GH} = \dfrac{AC}{FG}$.

(ii) In ∆ DCB and ∆ HGE,

⇒ ∠DBC = ∠HEG [∵ ∆ ABC ~ ∆ FEG]

⇒ ∠DCB = ∠HGE [∵ $\dfrac{∠ACB}{2} = \dfrac{∠FGE}{2}$]

∴ ∆ DCB ~ ∆ HGE (By A.A. axiom)

Hence, proved that ∆ DCB ~ ∆ HGE.

(iii) In ∆ DCA and ∆ HGF,

⇒ ∠DAC = ∠HFG [∆ ABC ~ ∆ FEG]

⇒ ∠ACD = ∠FGH [From (1)]

∴ ∆ DCA ~ ∆ HGF (By A.A. axiom)

Hence, proved that ∆ DCA ~ ∆ HGF.