__Instruction:__ Construct the point where the angle bisectors of the triangle are intersected.

__Goal:__ 2L 6E

Available tools:

- Move
- Point
- Line
- Circle
- Perpendicular Bisector
- Angle Bisector
- Intersect

__Pack:__ Beta

__Previous level:__ 2.1

__Next level:__ 2.3

## 2L solution[]

Given the triangle ABC

- Construct the angle bisector A1 of ∠BAC
- Construct the angle bisector A2 of ∠ACB, intersecting A1 at D
- Point at D

## 6E solution[]

Given the triangle ABC

- Note: the first two circles C1 and C2 must intersect with the specified sides. If they don't, either move vertex C, or choose another pair of vertices

- Construct the circle C1 with center A and radius AB, intersecting AC at D
- Construct the circle C2 with center B and radius AB, intersecting BC at E
- Construct the circle C3 with center D and radius AD, intersecting circle C2 at F
- Construct line AF
- Construct the circle C4 with center E and radius BE, intersecting circle C1 at G
- Construct line BG, intersecting AF at H
- Point at H

## Explanation[]

Solution 6E has you manually contruct the two angle bisectors. The first two circles has you construct two isosceles triangles ABD and ABE. Considering ABD with its base BD and the circle B, you can construct another circle with center D and the same radius and both circles would intersect along the perpendicular bisector of BD, which coincides with the angle bisector of ∠BAD since ABD is isosceles. Use the same reasoning by considering ABE with is base AE and the circle A and you can construct the perpendicular bisector of AE, which again coincides with the angle bisector of ∠ABE.