Problem
GEO-B3-M06-P002 Ratio on a Side
#2
★☆☆☆☆ Level 1 of 5
Cevian \(AD\) of triangle \(ABC\) meets \(BC\) at \(D\). Prove that \(\frac{BD}{DC}=\frac{AB\sin\angle BAD}{AC\sin\angle CAD}\).
Inspired by Prasolov trigonometric geometry method
C. Hint 1. Compare areas \(ABD\) and \(ACD\) in two ways.
D. Hint 2. First through bases on \(BC\), then through sides \(AB,AC\).
Since triangles \(ABD\) and \(ACD\) have altitudes to the same line \(BC\), \(\frac{S_{ABD}}{S_{ACD}}=\frac{BD}{DC}\).
Using the sine area formula, \(S_{ABD}=\frac12 AB\cdot AD\sin\angle BAD\), \(S_{ACD}=\frac12 AC\cdot AD\sin\angle CAD\). Dividing gives the formula.
This formula should become automatic.