Given: Point A is on the bisector of ∠PQR.

Prove: Point A is equidistant from the sides of ∠PQR.

Proof:
- QP ∥ QA ∥ QR
- QX ⊥ QP, and QY ⊥ QR
- QA is the bisector of ∠PQR.
- QX and QY are perpendiculars drawn from A to QP and QR, respectively.
- ∠QXA = ∠QYA = 90°: Perpendiculars create right angles.
- AX = AY: A is equidistant from the sides of ∠PQR, as it lies on the bisector.

Point A is equidistant from the sides PQ and QR of angle PQR when it is on the bisector of angle PQR. Using congruent triangles and the properties of bisectors, we've proven that the perpendicular segments from A to the sides PQ and QR are equal, hence confirming A's equal distance.

Explanation:

Let's consider our given that point A is on the bisector of ∠PQR. This means that ∠PQA = ∠AQR. From this we can say that triangle PQA and triangle AQR are similar by the Angle-Angle postulate (AA).

If point A is indeed equidistant from sides PQ and QR, it suggests that we have two segments AX and AY, where AY is perpendicular to side QR, and AX is perpendicular to PQ. Since ∠QXA=∠QYA=90∘, we know that PQA and AYQ are right triangles.

By our prior conclusion of ∠PQA = ∠AQR and considering ∠QAX = ∠YQA = 90∘, we can confirm that triangle PQA is congruent to triangle AQR by the Angle-Angle (AA) criterion for congruence (Note: In right-angled triangles, congruence is given by the AA criterion). Hence, the corresponding sides i.e. QX=QY making Point A equidistant from the sides of ∠PQR.

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