How to prove that the excentral triangle passes through the vertices of the original triangle?

Question

For a triangle ABC let's say I1, I2, I3 , are the three excentres opposite to angle A , B and C , respectively . Now if we join I1I2 , I2I3 , and I1I3 , how can we be sure that they will pass through vertex C , A and B respectively ? Please guide me to the solution .

Answer 1

I will denote rather by $I_A$, $I_B$, $I_C$ the three ex-centers. They are located each on one interior angle bisector, and two exterior angle bisectors of the angles of the given triangle $\Delta ABC$.

We compute now the angle $\widehat{I_BAI_C}$ as the sum of three angles, $$ \widehat{I_BAI_C} = \widehat{I_BAC} + \widehat{CAB}+ \widehat{BAI_C} = \frac 12(\pi-\hat A)+\hat A+\frac 12(\pi-\hat A)=\pi\ . $$ So the points $I_B$, $A$, $I_C$ are collinear.

$\square$