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Suppose we have a triangle $ABC$. Let us extend the sides $BA$ and $BC$. We will draw the incircle of this triangle.

- Here is the construction.
- Draw any two angle bisectors, say of angle $A$ and angle $B$
- Mark the intersection point $I$.
- Drop a perpendicular line from I to one of the sides, say $AC$ in this picture.
- Suppose the perpendicular intersects $AC$ at $E$.
- The incircle is drawn centred at $I$ and with radius $IE$

Suppose EI intersects the incircle at F.

Now let us draw the excircle.

To do that we will need the angle bisector of external angle A and external angle C. Suppose they intersect at $I_A$. Drop a perpendicular from $I_A $ to extended $BA$ or extended $BC$ or $AC$. In this picture we drop it on extended $BA$ Suppose $J$ is the point of intersection of extended $BA$ and the perpendicular.

Draw a circle centred at $I_A$ and radius $I_A J$. This is the excircle.

The incircle can be dilated or blown up with respect to point $B$ into the excircle. The center $I$ is sent to the center $I_A$ under dilation $FE$ which is perpendicular to $AC$ is sent to another segment perpendicular to $AC$ as angles are preserved under dilation

- What is the ratio of dilation?
- How can you rigorously show that the center $I$ goes to the center $I_A$ under this dilation?
- Where do the point $F$ go under this dilation?
- Show that $AN = CE$

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