The three angle bisectors of the internal angles of a triangle are concurrent. The angle bisectors that we want to construct are shown in green below.

Given triangle ABC, we want the angle bisectors to meet in a point.

Construct the angle bisector for angle ABC and angle ACB. This can be done using Euclid’s proposition I.9. Label the bisectors BF and CE respectively.

From point D, construct a line perpendicular to AB, AC, and CB. This can be done using Euclid’s proposition I.12. These perpendiculars are shown in pink.

We bisected angle ABC, we know angle ABD is equal to angle DBC. Similarly, since we bisected angle ACB, we know angle ACD is equal to angle DCB. We also know that perpendicular lines create right angles. This means since:

1. GD is perpendicular to AB, angle AGD and angle DGB are right

2. DI is perpendicular to BC, angle DIB and angle DIC are right angles

3. DH is perpendicular to AC, angle AHD and angle DHC are right.

Finally, since lines BD and CD are common, they are equal.

This means triangle EDB is congruent to triangle IDB by angle angle side (Euclid’s proposition I.26). Similarly, triangle HDC is congruent to triangle IDC by angle angle side too.

Since triangle EDB is congruent to triangle IDB, all corresponding sides and angles are equal. Namely, line DG is equal to line DI. Similarly, since triangle HDC is congruent to triangle IDC, all corresponding sides and angles are equal. Namely, line DI is equal to line DH. Hence, lines DG=DI=DH, so line DG is equal to line DH.

Now construct line AD.

Notice, angle AGD is equal to angle AHD as both are right angles (since AB is perpendicular to GD and AC is perpendicular to HD). We just found line DG is equal to DH. And AD is common. By the hypotenuse-leg theorem for right triangles, triangle ADG is congruent to triangle ADH.

Since triangle ADG is congruent to triangle ADH, we know all corresponding angles and segments are equal. Namely, angle GAD is equal to angle HAD. Since angle GAD is equal to HAD, angle BAC has been bisected.

Thus the angle bisectors meet in a point D.

Propositions of Euclid used:

Proposition I.9 To bisect a given rectilinear angle in half.

Proposition I.12 To draw an infinite straight line, from a given point which it is not on it, to draw a perpendicular straight.

Proposition I.26 If two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle to the remaining angle. [Commonly known as Angle Side Angle (ASA) and Angle Angle Side (AAS)]

For explanations of Euclid’s proofs, please see Euclid’s Elements Book I.

Proof of the Hypotenuse-Leg theorem for right triangles:

Given two triangles, ABC and DEF with line AB equal to line DE, line AC equal to line DF and angle ABC equal to angle DEF. Both angle ABC and angle DEF are right angles.

Since line AB is equal to line DE, we can use rigid motions to move triangle ABC so that AB lies on DE. Since the lines are equal, they will overlay each other perfectly. Now flip triangle ABC over line AB to get the following construction:

Notice line CF will be a straight line since both angle ABC and angle DEF are right and two right angles that are adjacent will form a straight line (proposition I.16). Looking at triangle ACF, we know it is isosceles since line AC is equal to line DF (which is now line AF). Hence angle ACF is equal to angle AFC (proposition I.5). Now we can use angle, angle, side to show that triangle ACB is congruent to triangle AFB (proposition I.26). Hence triangle ACB is congruent to triangle DFC.