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Angle bisector

16 viewed last edited 1 year ago
Anthony Schwartz
0
How do I show that the angle bisector of an angle in a triangle divides the opposite side in the same proportion as the adjoining sides?
Mahesh Godavarti
0
Essentially, given \triangle ABC , where AD is the angle bisector, we want to show \frac{AB}{AC} = \frac{BD}{BC} . We extend AD to E , so that BE \cong BD . Then, we have \triangle ADC \sim \triangle AEB using the AAA postulate. Reasons: 1. \angle DAC \cong \angle EAB (by definition) 2. \angle BEA \cong \angle CDA . Reasons: i) \angle ADC \cong \angle BDE (vertical angles) ii) \angle BDE \cong \angle BED = \angle BEC ( \triangle BDE is an isosceles triangle) Therefore, \frac{AB}{AC} = \frac{BE}{DC} = \frac{BD}{DC} .