Let [math]\triangle{ABC}[/math] have a right angle at [math]C[/math], and construct points [math]D[/math] and [math]E[/math] such that [math]BC=BD=BE[/math].
Note that [math]\angle AEC = 90 - \frac{ \angle CBE }{2} = \frac{\angle ABC}{2} = 90 - \angle BCD = \angle ACD[/math], so [math]\triangle{AEC}[/math] and [math]\triangle{ACD}[/math] are similar by [math]AA[/math].
Using this similarity, [math]\frac{AD}{AC}=\frac{AC}{AE}[/math], which rearranges to [math]AC^2=AE\cdot AD = (AB + BE)(AB - BD)=(AB + BC)(AB - BC)=AB^2-BC^2[/math], as desired.
You might ...
Không có nhận xét nào:
Đăng nhận xét