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Question

CD and GHare respectively the bisectors of ACB and EGF such that D and H lie on sidesAB and FE of ΔABC and ΔEFG

respectively. If ΔABC~ΔFEG,

Show that:

(i)CDGH=ACFG

(ii) ΔDCB~ΔHGE

(iii) ΔDCA~ΔHGF


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Solution

Similarity of triangle:

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.

To Prove

If ΔABC~ΔFEG,

(i) CDGH=ACFG
(ii)ΔDCB~ΔHGE
(iii) ΔDCA~ΔHGF

Proof

(i) We have,

ΔABC~ΔFEG.

Therefore,

A=F,B=E, and ACB=FGE

Now, since,

ACB=FGE

ACD=FGH (Angle bisector)

And, DCB=HGE

In ΔACDand ΔFGH,

A=FACD=FGH

By AAsimilarity criterion, we have,

ΔACD~ΔFGHCDGH=ACFG

(ii) In ΔDCB and ΔHGE,

Proved in part (i)

DCB=HGEB=E

ByAA similarity criterion,

ΔDCB~ΔHGE

(iii) In ΔDCA and ΔHGF,

Since, we have already proved,

ACD=FGHA=F

By AAsimilarity criterion,

ΔDCA~ΔHGF

Hence Proved


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