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) $$\dfrac{ CD}{GH}=\dfrac{AC}{FG}$$

(ii) ΔDCB ~ ΔHGE

(iii) ΔDCA ~ ΔHGF

Asked by Abhisek | 1 year ago |  88

##### Solution :-

(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)

$$\dfrac{CD}{GH}=\dfrac{AC}{FG}$$

(ii) In ΔDCB and ΔHGE,

ΔDCB ~ ΔHGE (AA similarity criterion)

(iii) In ΔDCA and ΔHGF,

ΔDCA ~ ΔHGF (AA similarity criterion)

Answered by Pragya Singh | 1 year ago

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