The side BC of ΔABC is produced to D. The bisectors of ∠ABC and ∠ACD meet at E. If AB = AC and ∠BEC = 35°, then the measure of ∠ABC will be:

55°

We have,

AB = AC

∠BEC = 35° 

= AB = AC

= ∠ACB = ∠ABC

We know,∠ACB + ∠ACD = 180°  (linear angles)

=∠ACB = 180° - ∠ACD

=∠ACB = 180° - 2b

= 2a = 180° - 2b

= 2a + 2b = 180°

= a + b = 90°

In ΔBEC, exterior angle equal to sum of exterior angles.

=∠EBC + ∠BEC = ∠ECD

= a + 35° = b

= b - a = 35°

Solving,

= b = 62.5° and a = 27.5°

∠ABC = 2x = 2 × 27.5 = 55°