Question

based on the figure to the right, explain why the following statement is false.
pq=ps
choose the correct answer below.
a. point p is the circumcenter of the triangle, so it is equidistant from the vertices of the triangle. ( overline{pq} perp overline{ab} ) and ( overline{ps} perp overline{ac} ) would both need to be true for ( pq = ps ) to be true.
b. point p is the circumcenter of the triangle, so it is equidistant from the sides of the triangle. ( pq ) would only equal ( ps ) if ( ps = pb ) and ( pq = pc ).
c. point p is the incenter of the triangle, so it is equidistant from the vertices of the triangle. ( pq ) would only equal ( ps ) if ( ps = pb ) and ( pq = pc ).
d. point p is the incenter of the triangle, so it is equidistant from the sides of the triangle. ( overline{pq} perp overline{ab} ) and ( overline{ps} perp overline{ac} ) would both need to be true for ( pq = ps ) to be true.

Explanation:

To determine why \( PQ = PS \) is false, we analyze the properties of triangle centers:

• Incenter (D) is equidistant from the triangle’s sides (perpendicular distances). For \( PQ = PS \), \( PQ \perp AB \) and \( PS \perp AC \) must hold (since incenter distances are perpendicular to sides). The diagram shows angle bisectors (not perpendiculars), so \( PQ \) and \( PS \) are not perpendicular to \( AB \) and \( AC \), violating the incenter’s distance property.

• Other options (A, B, C) misapply circumcenter (equidistant from vertices, not sides) or incenter logic. Only D correctly identifies \( P \) as incenter but explains \( PQ

eq PS \) because the perpendicularity condition fails.

Answer:

D. Point P is the incenter of the triangle, so it is equidistant from the sides of the triangle. \( \overline{PQ} \perp \overline{AB} \) and \( \overline{PS} \perp \overline{AC} \) would both need to be true for \( PQ = PS \) to be true.