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It is known that in the triangle ABC, the angle B is equal to twice the angle c, and the proof of AC = AB + BD is given
It is proved that AE = AB is intercepted on AC,
Because, ad bisector BAC
So ∠ bad = ∠ ead,
And ad is the public side,
So △ bad ≌ △ ead (SAS),
So BD = ed, ∠ AED = B,
In △ CDE, ∠ AED = ∠ EDC + ∠ C
And ∠ B = 2 ∠ C
So, EDC = EDC
So de = EC,
So AC = AE + EC = AB + de = AB + BD
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