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)`

In the given figure, CD and GH are respectively the bisectors of `angl

1.	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)`
Answer» `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.]