# Euclid IV-4. (pg 86)

- Author:
- Emily Crum

Explanation:
Let ABC be a triangle. Let angle ABC be bisected and angle BCG be bisected. Where there two bisectors meet, let it be called point D. Draw ED perpendicular to AB, draw DG perpendicular to AC, and draw DF perpendicular to BC. From points E,G, and F create a circle.
Because angle ABC was bisected, angle EBD = angle FBD. Also since ED is perp. to AB and FD is per. to BC, angles DEB and DFB are right angles and thus, angle DEB= angle DFB. Finally, BD=BD. Thus by AAS, triangle EDB is congruent to triangle FDB. Since they are congruent, this means that ED=FD.
Because angle ACB was bisected, angle GCD = angle FCD. Also since DG is perp. to AC and FD is per. to BC, angles DGC and DFC are right angles and thus, angle DGC= angle DFC. Finally, DC=DC. Thus by AAS, triangle DCF is congruent to triangle DCG Since they are congruent, this means that DG=FD.
Thus by transitivity, ED=DG=FD. Therefore circle EFG had been created with center D and radius FD.
(Furthermore, since ED, DG, and FD are all drawn perpendicular to their respective sides, this means that ED, DG, and FD are the shortest distance to those sides and thus any radii cannot pass through any of these 3 sides and only pts E, G, and F can touch these sides.)
Thus a circle has been inscribed in a triangle