Question

在△ABC中,∠C=2∠B,AD⊥BC于点D,求证:BD=AC+CD

Answer 1

在BD上取一点E,使得：DE = CD .

因为,△ACE中,AD⊥CE ,DE = CD ,

所以,△ACE是等腰三角形,AE = AC ,∠AEC = ∠C = 2∠B .

因为,△ABE中,∠BAE = ∠AEC - ∠B = ∠C - ∠B = ∠B ,

所以,△ABE是等腰三角形,BE = AE = AC .

可得：BD = BE+DE = AC+CD .