Question

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

Show that :

(i)
CD/GH=AC/FG

(ii)
triangle DCB ~ triangle HGE

(iii)
triangle DCA ~ triangle HGF.​

Answer 1

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 = ∠FGE

Since, ∠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)