As shown in the figure, in △ ABC, we prove: (1) if ad is the bisector of ∠ BAC, then s △ abd: s △ ACD = AB: AC; (2) let d be a point on BC to connect ad, if s △ abd: s △ ACD = AB: AC, then ad is the bisector of ∠ BAC

(1) It is proved that if ah ⊥ BC is made through a and CE ∥ AB is made through C and ad extension line is made through e, then ∠ e = ∠ bad, ∥ ad bisection ∠ BAC, ∥ CAD = ∠ bad, ∥ e = ∠ CAD, ∥ AC = CE, ∥ CE ∥ AB, ∥ ECD ∥ abd, ∥ bdcd = abce, ∥ bdcd = ABAC, ∥ s △ abd: s △ ACD = (12 × BD × ah

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