Question

In trapezium ABCD ,ABllDC ,E AND F ARE points on non parallel sides AD and BC respectively such that EFllAB .Show that
$\frac{ae}{ed} = \frac{bf}{fc}$

Answer 1

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EF II DC = AD II EF and AD II DC

Answer 2

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The answer of u r question....

Ans:

Given,

AB ll DC AND EF ll AB

= EF ll DC (lines parallel to the same line are parallel to each other..)

In triangle ADC, EF ll DC

so,

$\frac{ae}{ed} = \frac{ag}{gc} by \: bbt$

Similarly,

In triangle CAB ,GF ll AB

$\frac{cg}{ga} = \frac{cf}{fb} (by \: bbt)$
from (1) and (2)

$\frac{ae}{ed} = \frac{bf}{fc}$

Hence proved...

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