Angle Bisector and Equidistance

An animated proof demonstrating that any point on an angle bisector is equidistant from the two sides of the angle, and conversely, any point equidistant from the two sides lies on the angle bisector.

Keywords: angle bisector, equidistant, congruent triangles, AAS congruence, HL theorem

Prerequisites: triangle-congruence · Difficulty: beginner

This proof establishes a beautiful two-way relationship:

1. Forward: If a point lies on an angle bisector, it is equidistant from both sides
2. Converse: If a point is equidistant from both sides, it lies on the angle bisector

Together, these show that the angle bisector is exactly the set of all points equidistant from the two sides.

Why This Matters

This property is the foundation for the incenter of a triangle. Since the incenter lies on all three angle bisectors, it must be equidistant from all three sides — which is why it's the center of the inscribed circle!

The Proof

Step 1: Part 1: Setup

We start with angle $\angle BAC$ and its angle bisector.

Let $P$ be any point on the bisector. We'll prove that $P$ is equidistant from both sides — that is, the perpendicular distances $PD$ and $PE$ are equal.

Step 2: Proving $PD = PE$ by AAS

Form triangles $\triangle PAD$ and $\triangle PAE$. We prove they're congruent:

• $\angle PAD = \angle PAE$ (P is on the bisector)
• $\angle PDA = \angle PEA = 90°$ (perpendiculars)
• $PA = PA$ (common side)

By AAS congruence, $\triangle PAD \cong \triangle PAE$, so $PD = PE$.

Step 3: Part 2: The Converse

Now we prove the reverse: if a point $P$ is equidistant from both sides ($PD = PE$), then $P$ must lie on the angle bisector.

Step 4: Proving $P$ is on the Bisector by HL

Form right triangles $\triangle PAD$ and $\triangle PAE$. We prove congruence:

• $PD = PE$ (given)
• $PA = PA$ (common hypotenuse)
• Both have right angles

By the Hypotenuse-Leg theorem, $\triangle PAD \cong \triangle PAE$.

Therefore $\angle PAD = \angle PAE$, so ray $AP$ bisects the angle — $P$ is on the bisector!

Step 5: Conclusion

We've proven both directions:

• Forward: Points on the bisector are equidistant from both sides
• Converse: Points equidistant from both sides lie on the bisector

The angle bisector is the locus of all points equidistant from the two sides!

Conclusion

We've proven both directions:

• Forward: Points on the bisector are equidistant from both sides (by AAS congruence)
• Converse: Points equidistant from both sides lie on the bisector (by HL theorem)

The angle bisector is exactly the locus of points equidistant from the two sides. This is why angle bisectors are so important for finding the incenter!