Friday, 9 January 2015

A cute geometry problem about angle trisectors.

Here is a cute geometry problem I saw some time ago. I know the solution, I just wanted to share ;-) (Please, don't be mad at me.)



Consider an acute triangle $\triangle ABC$. Let $AP$, $AQ$ and $BP$,$BQ$ be the angle trisectors as shown on the picture below. Prove that $|\angle APQ| = |\angle QPB|$.



$\hspace{60pt}$ enter image description here



Edit: There does exist one very simple and elegant solution, so don't be stumbled if you happen to guess/derive it.

No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...