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Question
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
- `("CD")/("GH") = ("AC")/("FG")`
- ΔDCB ~ ΔHGE
- ΔDCA ~ ΔHGF
Solution
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Given:
(i) ΔABC ∼ ΔFEG
∠A = ∠F ...(1)
∠B = ∠E ...(2)
∠C = ∠G ...(3)
The sides opposite to equal angles are essentially equal.
`"CD"/"GH" = "AC"/"FG"`
(ii) In ΔDCB and ΔHGE
In equation (2),
∠B = ∠E
In equation (3),
∠C = ∠G
∠BCD = ∠EGH
AA by similarity criterion
∆DCB ∼ ∆HGE
(iii) In ΔDCA and ΔHGF
In Equation (1)
∠A = ∠F
In equation (3),
∠C = ∠G
∠ACD = ∠FGH
AA by similarity criterion
∆DCA ∼ ∆HGF
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