三角形ABC中角C等于2角B,AD垂直BC于D,试说明BD等于AC+CD

作△ACD以DA为轴,翻折,得△AC'D
∴∠ACD=∠AC'D,CD=C'D
∵∠BCA=2∠CBA
而∠DC'A=∠CBA+∠C'AB
2∠CBA=∠CBA+∠C'AB
∠CBA=∠C'AB
∴C'B=C'A(等角对等边)
∴BD=C'D+C'B
=CD+CA