Question

point p is the incenter of δrst. which must be true?
overline{cp}congoverline{rp}
overline{bp}congoverline{pc}
angle psccongangle ptc
angle bpscongangle apt

Explanation:

Step1: Recall the incenter property

The incenter \( P \) of a triangle is the intersection of angle bisectors. So, \( SP \) bisects \( \angle RST \), \( TP \) bisects \( \angle RTS \), and \( RP \) bisects \( \angle SRT \). Also, the distances from the incenter to the sides (the perpendiculars) are equal, but we focus on angle bisectors here.

Step2: Analyze each option

• Option 1: \( \overline{CP} \cong \overline{RP} \): \( CP \) is a perpendicular distance (inradius - like) and \( RP \) is a segment from incenter to vertex. No reason for them to be congruent. Eliminate.
• Option 2: \( \overline{BP} \cong \overline{PC} \): \( BP \) and \( PC \) are perpendiculars to different sides (unless triangle is isoceles, not given). No guarantee. Eliminate.
• Option 3: \( \angle PSC \cong \angle PTC \): \( \angle PSC \) is related to \( SP \) and \( \angle PTC \) to \( TP \). No direct relation from incenter. Eliminate.
• Option 4: \( \angle BPS \cong \angle APT \): Since \( SP \) bisects \( \angle RST \) and \( TP \) bisects \( \angle RTS \), and \( \angle BPS \) and \( \angle APT \) are vertical angles or related by angle bisector properties? Wait, actually, \( SP \) bisects \( \angle RST \), so \( \angle BPS = \angle SPC \) (wait, no, \( PB \perp SR \), \( PC \perp ST \), so \( SP \) bisects \( \angle RST \), so \( \angle BPS = \angle CPS \). Similarly, \( TP \) bisects \( \angle RTS \), so \( \angle APT = \angle CPT \). But also, \( \angle BPS \) and \( \angle APT \): Wait, actually, the key is that \( \angle BPS \) and \( \angle APT \) are equal because of the angle bisector and the right angles. Wait, more accurately, since \( PB \perp SR \), \( PA \perp RT \), \( PC \perp ST \), and \( P \) is incenter (angle bisectors), so \( \angle BPS \) and \( \angle APT \) are equal because \( SP \) and \( TP \) are angle bisectors, and the angles formed with the perpendiculars lead to congruent angles. Alternatively, using the fact that the incenter creates angle bisectors, and the angles \( \angle BPS \) and \( \angle APT \) are equal due to the symmetry of angle bisectors and right angles. So this is the correct one.

Answer:

\(\boldsymbol{\angle BPS \cong \angle APT}\) (the fourth option, i.e., the option with \(\angle BPS \cong \angle APT\))