Question

CD and GH are respectively the bisectors of ΔACB and ΔEGF, such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, show that:
(i) $\frac{CD}{GH}=\frac{AC}{FG}$
(ii) ΔDCBΔHGE
(iii) ΔDCAΔHGF

Answer 1

Given: ΔABC ~ ΔFEG
CD and GH are respectively the bisectors of ΔACB and ΔEGF, such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively.

To Prove:

1. $\frac{CD}{GH}=\frac{AC}{FG}$
2. ΔDCB~ΔHGE
3. ΔDCA~ΔHGF

Proof:

(i) It is given that ∆ABC ∼ ∆FEG.
∴ $\angle$A = $\angle$F, ∠B = ∠E, and $\angle$ACB = $\angle$FGE
∴ $\angle$ACD = $\angle$FGH (Angle bisector)
And, $\angle$DCB = HGE (Angle bisector)
In ∆ACD and ∆FGH, $\angle$A = $\angle$F (Proved above)
$\angle$ACD = $\angle$FGH (Proved above)
∴ ∆ACD ∼ ∆FGH (By AA similarity criterion)
⇒$\frac{CD}{GH}=\frac{AC}{FG}$

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(ii) In ∆DCB and ∆HGE, $\angle$DCB = $\angle$HGE (Proved above)
$\angle$B = $\angle$E (Proved above)
∴ ∆DCB ∼ ∆HGE (By AA similarity criterion)

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(iii) In ∆DCA and ∆HGF, $\angle$ACD = $\angle$FGH (Proved above)
$\angle$A = $\angle$F (Proved above)
∴ ∆DCA ∼ ∆HGF (By AA similarity criterion)