Exercise 6.3 Question no. 10

1.	Exercise 6.3 Question no. 10
Answer» 10. 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\xa0~\xa0ΔHGE(iii) ΔDCA ~ ΔHGFSolution:Given, 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.(i) From the given condition,ΔABC ~ ΔFEG.∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGESince, ∠ACB = ∠FGE∴ ∠ACD = ∠FGH (Angle bisector)And, ∠DCB = ∠HGE (Angle bisector)In ΔACD and ΔFGH,∠A = ∠F∠ACD = ∠FGH∴ ΔACD ~ ΔFGH (AA similarity criterion)⇒CD/GH = AC/FG(ii) In ΔDCB and ΔHGE,∠DCB = ∠HGE (Already proved)∠B = ∠E (Already proved)∴ ΔDCB ~ ΔHGE (AA similarity criterion)(iii) In ΔDCA and ΔHGF,∠ACD = ∠FGH (Already proved)∠A = ∠F (Already proved)∴ ΔDCA ~ ΔHGF (AA similarity criterion)

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