In equilateral Δ ABC, D and E are points on the sides AB…

In equilateral $\Delta$ ABC, D and E are points on the sides AB and AC, respectively, such that AD = CE. BE and CD intersect at F. The measure (in degrees) of $\angle$ CFB is:

Show answer & explanation

Correct answer: A

Explanation

The correct answer is option 1 i.e. 120º.

In $\Delta$ ADC and $\Delta$ BEC ,

AD = EC (given) , $\angle$ BAE = BCE and AC = BC.

So $\Delta$ ADC and $\Delta$ BEC are congruent to each other.

So, $\angle$ EBC and $\angle$ DCA are equal (let $\theta$ )

$\angle$ FCB = 60 - $\theta$ .

By using the triangle sum property in $\Delta$ BFC we get,

$\angle$ BFC + $\angle$ FCB + $\angle$ FBC = 180°

$\angle$ BFC = 180 - $\theta$ - 60 + $\theta$

$\angle$ BFC = 120°

SSC CGL 2022 PYQ — Question 23