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Babita Dave
Asked: 2022-11-01T11:07:29+05:30 In:

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)`
image

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)`
image
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  1. `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.]

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