![左鍵: 點擊縮放; 右鍵: 觀看原圖](http://yll.loxa.edu.tw/jpg/a/03062109014868.gif)
ABC is a triangle, D lies on BC, such that ∠BAD=∠CAD, then AB/BD=AC/CD. (Interior angle bisector ratio theorem)
Proof
![左鍵: 點擊縮放; 右鍵: 觀看原圖](http://yll.loxa.edu.tw/jpg/a/03062109133566.gif)
E lies on AB, such AC//ED. Join DE.
DE//CA (Given)
∴BD/DC=BE/EA (Equal ratios)
∠EAD=∠CAD (Given)
∴∠EDA=∠CAD (Alternate angles, DE//CA)
∴∠EDA=∠EAD
∴ED=EA (Side opposite equal angles)
∴BD/DC=BE/ED
∠DBE=∠CBA (Common angle)
∠BED=∠BAC (Corresponding angles, DE//CA)
∠EDB=∠ACB (Corresponding angles, DE//CA)
∴△ABC∼△EBD (AAA)
∴BE/ED=BA/AC (Corresponding sides, similar triangles)
∴BD/DC=BA/AC
∴AB/BD=AC/CD