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) ΔDCB~ΔHGE
(iii) ΔDCA~ΔHGF

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)