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When I was in the eighth grade of mathematics in a famous school.
△ABC, AB=AC, BD is AC high, ∠ABD=50.

In the right triangle ABD

Available < a = 90-50 = 40

AB = AC

∴∠ABC=∠ACB

That is ∠ ABC =∠ ACB = (180-40)/2 = 70.

Yes, there is another possibility.

AB=AC, BD is AC high, ∠DBA=50.

In the right triangle ABD

Available D = 90-50 = 40

∴∠BAC= 180-40= 140

AB = AC

∴∠ABC=∠ACB

That is ∠ ABC =∠ ACB = (180-140)/2 = 20.