CD and OH are respectively the bisectors of ∠ACB and ∠EGF such tha

1.	CD and OH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively.If ∆ABC ~ ∆EFG, show that (i) \(\frac{CD}{GH} = \frac{AC}{FG}\)(ii) ∆DCB ~ ∆HGE (iii) ∆DCA ~ ∆HGF
Answer» Data : CD and OH are respectively the bisectors of ∠ACB and ∠EGF such that D and H 11e on sides AB and FE of ∆ABC and ∆EFG respectively. ∆ABC ~ ∆EFG To Prove: (i) \(\frac{CD}{GH} = \frac{AC}{FG}\) (ii) ∆DCB ~ ∆HGE (iii) ∆DCA ~ ∆HGF Proof: ∆ABC ~ ∆EFG (data given) ∴ Their corresponding sides are in proportion. ∴ \(\frac{AB}{EF} = \frac{BC}{FG} = \frac{AC}{EG}\) ∠B = ∠F, ∠A = ∠E, ∠C = ∠G. (i) In ∆ADC and ∆EHG, ∠A = ∠E . ∠ACD = ∠EGH ∴ Their sides are in proportion. ∴ \(\frac{CD}{GH} = \frac{AC}{FG}\) (ii) In ∆DCB and ∆HGE. ∴ \(\frac{CD}{GH} = \frac{AC}{FG}\) ∴ ∆DCB ~ ∆HGE (iii) In ∆DCA and ∆HGE, ∴ \(\frac{DC}{GH} = \frac{AD}{EH} = \frac{AC}{EG}\) ∴ ∆DCA ~ ∆HGE

CD and OH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively.If ∆ABC ~ ∆EFG, show that (i) \(\frac{CD}{GH} = \frac{AC}{FG}\)(ii) ∆DCB ~ ∆HGE (iii) ∆DCA ~ ∆HGF

Answer»

Data : CD and OH are respectively the bisectors of ∠ACB and ∠EGF such that D and H 11e on sides AB and FE of ∆ABC and ∆EFG respectively.

∆ABC ~ ∆EFG

To Prove:

(i) \(\frac{CD}{GH} = \frac{AC}{FG}\)

(ii) ∆DCB ~ ∆HGE

(iii) ∆DCA ~ ∆HGF

Proof: ∆ABC ~ ∆EFG (data given)

∴ Their corresponding sides are in proportion.

∴ \(\frac{AB}{EF} = \frac{BC}{FG} = \frac{AC}{EG}\)

∠B = ∠F, ∠A = ∠E, ∠C = ∠G.

(i) In ∆ADC and ∆EHG,

∠A = ∠E . ∠ACD = ∠EGH

∴ Their sides are in proportion.

∴ \(\frac{CD}{GH} = \frac{AC}{FG}\)

(ii) In ∆DCB and ∆HGE.

∴ \(\frac{CD}{GH} = \frac{AC}{FG}\)

∴ ∆DCB ~ ∆HGE

(iii) In ∆DCA and ∆HGE,

∴ \(\frac{DC}{GH} = \frac{AD}{EH} = \frac{AC}{EG}\)

∴ ∆DCA ~ ∆HGE