Question

the angle bisectors of △efg are (overline{es}), (overline{fs}), and (overline{gs}). they meet at a single point ( s ). (in other words, ( s ) is the incenter of ( \triangle efg ).) suppose ( qs = 18 ), ( gs = 23 ), ( mangle qgr = 44^circ ), and ( mangle pes = 52^circ ). find the following measures. note that the figure is not drawn to scale.
( mangle per = square^circ )
( mangle pfs = square^circ )
( ps = square )

Explanation:

Step1: Find \( PS \)

Since \( S \) is the incenter, it is equidistant from all sides of the triangle. So \( PS = QS \). Given \( QS = 18 \), then \( PS = 18 \).

Step2: Find \( m\angle PER \)

\( ES \) is an angle bisector, and \( \angle PES = 52^\circ \). So \( \angle PER = 2\times\angle PES = 2\times52^\circ = 104^\circ \)? Wait, no, wait. Wait, \( \angle PES \) and \( \angle RES \) are equal because \( ES \) is the angle bisector. Wait, actually, \( \angle PER \) is a straight line? No, wait, \( PR \) and \( PQ \) are perpendiculars. Wait, no, \( S \) is the incenter, so \( ES \) bisects \( \angle E \). So \( \angle PES = \angle RES = 52^\circ \), so \( \angle PER = \angle PES + \angle RES = 52^\circ + 52^\circ = 104^\circ \)? Wait, no, maybe I messed up. Wait, \( \angle QGR = 44^\circ \), \( GS \) bisects \( \angle G \), so \( \angle EGF \) is related. Wait, let's correct.

Wait, \( S \) is the incenter, so \( ES \) bisects \( \angle E \), \( FS \) bisects \( \angle F \), \( GS \) bisects \( \angle G \).

Given \( \angle QGR = 44^\circ \), \( GS \) bisects \( \angle G \), so \( \angle EGF = 2\times44^\circ = 88^\circ \).

Given \( \angle PES = 52^\circ \), \( ES \) bisects \( \angle E \), so \( \angle E = 2\times52^\circ = 104^\circ \).

Then in triangle \( EFG \), angles sum to \( 180^\circ \), so \( \angle F = 180^\circ - \angle E - \angle G = 180^\circ - 104^\circ - 88^\circ = -12^\circ \)? No, that can't be. Wait, I must have messed up \( \angle PER \).

Wait, \( \angle PER \): \( PR \perp EG \), \( PQ \perp EF \), so \( \angle EPS = \angle EQS = 90^\circ \). \( ES \) is the angle bisector, so \( \angle PES = \angle QES = 52^\circ \). Then \( \angle PER \): Wait, \( \angle PER \) is at \( E \), between \( PR \) and \( ER \). Wait, maybe \( \angle PER \) is \( 2\times\angle PES \)? Wait, no, \( \angle PES = 52^\circ \), so \( \angle PER = 2\times52^\circ = 104^\circ \)? But then angle sum would be wrong. Wait, maybe \( \angle QGR = 44^\circ \) is \( \angle G \) bisected. So \( \angle G = 2\times44^\circ = 88^\circ \), \( \angle E = 2\times52^\circ = 104^\circ \), then \( \angle F = 180 - 88 - 104 = -12 \), which is impossible. So I must have messed up \( \angle PER \).

Wait, maybe \( \angle PER \) is \( \angle PES \times 2 \)? No, wait, \( \angle PES = 52^\circ \), so \( \angle PER = 52^\circ \times 2 = 104^\circ \), but then angle \( F \) would be negative, which is wrong. So maybe \( \angle PES \) is half of \( \angle E \), so \( \angle E = 2\times52 = 104 \), \( \angle G = 2\times44 = 88 \), then \( \angle F = 180 - 104 - 88 = -12 \), which is impossible. So I must have made a mistake. Wait, maybe \( \angle QGR = 44^\circ \) is \( \angle G \), not the bisected angle. Wait, the problem says \( m\angle QGR = 44^\circ \), \( GS \) is the angle bisector, so \( \angle QGR \) is half of \( \angle G \), so \( \angle G = 2\times44 = 88 \). \( \angle PES = 52^\circ \), \( ES \) is angle bisector, so \( \angle E = 2\times52 = 104 \). Then \( \angle F = 180 - 88 - 104 = -12 \), which is impossible. So I must have messed up \( \angle PER \).

Wait, maybe \( \angle PER \) is \( \angle PES \), no. Wait, \( \angle PER \): \( PR \) is perpendicular to \( EG \), \( PQ \) is perpendicular to \( EF \), so \( \angle EPS = \angle ERS = 90^\circ \). \( ES \) is the angle bisector, so \( \angle PES = \angle RES = 52^\circ \), so \( \angle PER = \angle PES + \angle RES = 52 + 52 = 104^\circ \). But then angle \( F \) is negative, which is wrong. So maybe the problem has \( \angle QGR = 44^\circ \) as \( \angle G \), not the bisected angle. Wait, the problem says " \( m\angle QGR = 44^\circ \) ", \( QGR \): \( Q \) is on \( FG \), \( R \) is on \( EG \), \( GR \perp EG \), \( GQ \perp FG \)? No, \( R \) and \( Q \) are feet of perpendiculars. So \( \angle QGR \) is \( \angle G \) bisected, so \( \angle G = 2\times44 = 88 \), \( \angle E = 2\times52 = 104 \), \( \angle F = 180 - 88 - 104 = -12 \), which is impossible. So I must have made a mistake in \( \angle PER \).

Wait, maybe \( \angle PER \) is \( \angle PES \), no. Wait, let's check \( PS \): since \( S \) is incenter, \( PS = QS = 18 \), that's correct.

Now \( m\angle PFS \): \( FS \) bisects \( \angle F \). We know \( \angle E = 104^\circ \), \( \angle G = 88^\circ \), so \( \angle F = 180 - 104 - 88 = -12^\circ \), which is impossible. So maybe \( \angle QGR = 44^\circ \) is \( \angle G \), not the bisected angle. So \( \angle G = 44^\circ \), then \( \angle E = 2\times52 = 104^\circ \), \( \angle F = 180 - 44 - 104 = 32^\circ \), so \( \angle PFS = \frac{1}{2}\angle F = 16^\circ \). Ah, that makes sense! So I misread \( \angle QGR \): \( \angle QGR = 44^\circ \) is \( \angle G \), not the bisected angle. So \( GS \) bisects \( \angle G \), so \( \angle G = 44^\circ \), so \( \angle F = 180 - 104 - 44 = 32^\circ \), so \( \angle PFS = \frac{1}{2}\times32 = 16^\circ \).

And \( \angle PER \): \( ES \) bisects \( \angle E \), \( \angle PES = 52^\circ \), so \( \angle E = 2\times52 = 104^\circ \), so \( \angle PER = 104^\circ \)? Wait, no, \( \angle PER \) is \( \angle E \), which is \( 104^\circ \).

Wait, let's re-express:

1. \( PS \): incenter is equidistant from sides, so \( PS = QS = 18 \).

2. \( m\angle PER \): \( ES \) bisects \( \angle E \), \( \angle PES = 52^\circ \), so \( \angle E = 2\times52^\circ = 104^\circ \), so \( m\angle PER = 104^\circ \).

3. \( m\angle PFS \): \( \angle G = 44^\circ \) (since \( \angle QGR = 44^\circ \) is \( \angle G \)), \( \angle E = 104^\circ \), so \( \angle F = 180 - 44 - 104 = 32^\circ \), \( FS \) bisects \( \angle F \), so \( m\angle PFS = \frac{1}{2}\times32^\circ = 16^\circ \).

Yes, that works. So:

- \( PS = 18 \)

- \( m\angle PER = 104^\circ \)

- \( m\angle PFS = 16^\circ \)

Answer:

\( PS = \boxed{18} \)

\( m\angle PER = \boxed{104^\circ} \)

\( m\angle PFS = \boxed{16^\circ} \)