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DSSSB TGT Maths Male Subject Concerned- 23 Sep 2018 Shift 1

Option 1 : 1/3

**Concept:**

Ceva's Theorem

According to this theorem, if AD, BE, CF are concurrent lines meeting at the point O of a Δ ABC then

\(\frac{{AF}}{{FB}} × \frac{{BD}}{{DC}} × \frac{{CE}}{{EA}} = 1\)

__Given:__

In ΔABC

D, E, F be the points on lines BC, CA, and AB respectively such that lines AD, BE, and CF are concurrent.

\(\dfrac{AF}{FB} = \dfrac{1}{2},\ \dfrac{BD}{DC} = \dfrac{2}{3}\)

__Calculation:__

In ΔABC

Using the result

__\(\frac{{AF}}{{FB}} × \frac{{BD}}{{DC}} × \frac{{EC}}{{EA}} = 1\)__

\(\frac{{1}}{{2}} × \frac{{2}}{{3}} × \frac{{EC}}{{EA}} = 1\)

\(\frac{{EA}}{{EC}} =\frac{1}{3}\)