In the given figure, CD and GH are respectively the bisectors of `angle ACB and angle FGE "of" Delta ABC and Delta EFG` respectively . If `Delta ABC~ Delta FEG`, prove that :
(a) `Delta ADC~ Delta FHG " " (b) Delta BCD ~ Delta EGH " " ( c ) (CD)/(GH)=(AC)/(FG)`

Explanation

`Delta ABC~ Delta FEG` (given)
`:. angle ACB= angle FGE` [ corresponding angles of similar triangles are equal]
`rArr (1)/(2) angle ACD=(1)/(2) angle FGE`.
(a) In `Delta ADC and Delta FHG`, we have
`angle DAC= angle HFG [ :. angle A = angle F "since" Delta ABC~ Delta FEG`]
and `angle ACD = angle FGH` [ proved above]
`:. Delta ADC~ Delta FHG` [ by AA- similarity]
(b) In `Delta BCD and Delta EGH`, we have
`angle DBC= angle HEG [ :. angle B= angle E "since" Delta ABC~ Delta FEG]`
`and angle DCB= angle HGE` [ proved above]
`:. Delta BCD~ Delta EGH`.
(c) We have
`Delta ADC~ Delta FHG` [ proved above]
And so, `(CD)/(GH)=(AC)/(FG)` [ crossponding sides of similar triangles are proportional.]