Question : In equilateral $\triangle $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:
Option 1: 120°
Option 2: 135°
Option 3: 125°
Option 4: 105°
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Correct Answer: 120°
Solution :
In $\triangle CBE$ and $ \triangle ACD$ AD = CE (Given) $\angle $B = $ \angle $C (each angle is 60º) BC = AC (sides of an equilateral triangle) $\triangle CBE \cong \triangle ACD$ [SAS congruency] So, the three angles of these two triangles are the same, Let $\angle $EBC = z then $\angle $ACD = z Now, $\angle $BEC = 180° – (60° + z) = 120° – z In $\triangle $ECF $\angle $CFB = (120° – z) + z (The exterior angle is always equal to the sum of the interior opposite angle) $\angle $CFB = 120° Hence, the correct answer is 120°.
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