As shown in the figure, ab = AC, ∠ BAC = 120 ° in △ ABC, ad ⊥ AC intersects BC at point D, and BC = 3aD

As shown in the figure, ab = AC, ∠ BAC = 120 ° in △ ABC, ad ⊥ AC intersects BC at point D, and BC = 3aD

It is proved that in △ ABC, ∵ AB = AC, ∵ BAC = 120 °, and ∵ ad ⊥ AC, ∵ DAC = 90 °, and ∵ C = 30 °, CD = 2ad, ∵ bad = ∵ B = 30 °, ad = dB, ∵ BC = CD + BD = AD + DC = AD + 2ad = 3aD