Question 10:
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ∼ ∆FEG, Show that:
(i) CD/GH = AC/FG
(ii) ∆DCB ∼ ∆HGE
(iii) ∆DCA ∼ ∆HGF
Answer:
It is given that ∆ABC ∼ ∆FEG.
∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE
∠ACB = ∠FGE
∴ ∠ACD = ∠FGH (Angle bisector)
And, ∠DCB = ∠HGE (Angle bisector)
In ∆ACD and ∆FGH,
∠A = ∠F (Proved above)
∠ACD = ∠FGH (Proved above)
∴∆ACD ∼ ∆FGH (By AA similarity criterion)
⇒ CD/GH = AC/FG
In ∆DCB and ∆HGE,
∠DCB = ∠HGE (Proved above)
∠B = ∠E (Proved above)
∴ ∆DCB ∼ ∆HGE (By AA similarity criterion)
In ∆DCA and ∆HGF,
∠ACD = ∠FGH (Proved above)
∠A = ∠F (Proved above)
∴ ∆DCA ∼ ∆HGF (By AA similarity criterion)
Latest Govt Job & Exam Updates: