in the given fig DE//BC and AD:DB=7:5 find ar(DFE)/ar(CFB)
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Step-by-step explanation:
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In △ABC, we have
DE||BC
⇒ ∠ADE=∠ABC and ∠AED=∠ACB [Corresponding angles]
Thus, in triangles ADE and ABC, we have
∠A=∠A [Common]
∠ADE=∠ABC
and, ∠AED=∠ACB
∴ △AED∼△ABC [By AAA similarity]
⇒ ABAD=BCDE
We have,
DBAD=45
⇒ ADDB=54
⇒ ADDB+1=54+1
⇒ ADDB+AD=59
⇒ ADAB=59⇒ABAD=95
∴
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