Question

etermining unknown measures
chords \\(\overline{de}\\), \\(\overline{ef}\\), and \\(\overline{fg}\\) are congruent.
determine the following measures.
\\(m\overarc{fg} = \square\\)
\\(m\angle ecf = \square\\)

Explanation:

Step1: Find \( m\overarc{FG} \)

Congruent chords subtend congruent arcs. Since \( \overline{DE} \), \( \overline{EF} \), and \( \overline{FG} \) are congruent, their subtended arcs are congruent. Wait, actually, looking at the central angles: Wait, no, the angle \( \angle ECF \) – Wait, first, the sum of central angles in a circle is \( 360^\circ \), but maybe we can find the measure of \( \overarc{FG} \) by the central angle. Wait, the chord \( FG \) and \( EF \)? Wait, no, the given angle is \( \angle EFC = 56.5^\circ \)? Wait, no, the diagram: \( C \) is the center? Wait, \( E, F, G \) are on the circle, \( C \) is the center. So \( CE, CF, CG \) are radii. Chords \( DE, EF, FG \) are congruent. So the arcs \( DE, EF, FG \) are congruent? Wait, no, the central angles for congruent chords are equal. Wait, first, let's find the measure of \( \overarc{FG} \). Wait, the sum of central angles: Let's see, the central angles around \( C \): we have angles \( 67^\circ \), \( 75^\circ \), and we need to find \( \angle ECF \) and \( \angle FCG \). Wait, but chords \( DE, EF, FG \) are congruent, so their subtended central angles are equal? Wait, no, \( DE \) and \( EF \) are congruent (marked with same tick), \( EF \) and \( FG \) are congruent (marked with same tick). So \( DE \cong EF \cong FG \), so their central angles \( \angle DCE \), \( \angle ECF \), \( \angle FCG \) are equal? Wait, no, wait the diagram: \( DE \) is congruent to \( EF \) (tick marks), \( EF \) is congruent to \( FG \) (tick marks). So \( DE \cong EF \cong FG \), so their central angles are equal. Wait, but we have some angles: \( \angle EFC = 56.5^\circ \)? Wait, no, \( \angle EFC \) is a triangle angle. Wait, \( CE = CF \) (radii), so triangle \( ECF \) is isoceles? Wait, no, \( CE = CF \), so \( \angle CEF = \angle CFE = 56.5^\circ \). Then, in triangle \( ECF \), the sum of angles is \( 180^\circ \), so \( \angle ECF = 180 - 56.5 - 56.5 = 67^\circ \)? Wait, that's one of the given angles. Wait, maybe I messed up. Wait, the problem is to find \( m\overarc{FG} \) and \( m\angle ECF \). Let's start with \( m\overarc{FG} \). Wait, congruent chords subtend congruent arcs. So if \( EF \) and \( FG \) are congruent, then \( \overarc{EF} \cong \overarc{FG} \). Wait, but what's the measure of \( \overarc{EF} \)? Wait, the central angle for \( \overarc{EF} \) is \( \angle ECF \). Wait, but maybe the key is that the central angle for a chord is equal to the measure of its subtended arc. Wait, let's check the sum of central angles. The total around \( C \) is \( 360^\circ \), but maybe we have a semicircle? No, the diagram is a circle. Wait, maybe the arcs \( DE, EF, FG \) are congruent, so their central angles are equal. Wait, but we have angles \( 67^\circ \) and \( 75^\circ \). Wait, no, let's look at the triangle \( ECF \): \( CE = CF \) (radii), so it's isoceles with \( \angle CFE = 56.5^\circ \), so \( \angle CEF = 56.5^\circ \), so \( \angle ECF = 180 - 56.5 - 56.5 = 67^\circ \). So \( \angle ECF = 67^\circ \), which is the central angle for arc \( EF \). Then, since \( EF \cong FG \) (chords), their central angles \( \angle ECF \) and \( \angle FCG \) are equal? Wait, no, \( EF \) and \( FG \) are congruent, so \( \angle ECF = \angle FCG \)? Wait, but then \( m\overarc{FG} = m\overarc{EF} = 67^\circ \)? Wait, but the angle given is \( 56.5^\circ \) at \( F \). Wait, maybe I made a mistake. Wait, no, the angle at \( F \) is \( \angle EFC = 56.5^\circ \), which is an inscribed angle? No, \( F \) is on the circle, \( C \) is the center, so \( CF \) is a radius, \( EF \) is a chord. So triangle \( ECF \) has \( CE = CF \) (radii), so it's isoceles with base \( EF \). So the base angles are \( \angle CEF \) and \( \angle CFE \), each \( 56.5^\circ \), so the vertex angle \( \angle ECF = 180 - 2 \times 56.5 = 67^\circ \). So \( \angle ECF = 67^\circ \), which is the central angle for arc \( EF \). Then, since chord \( FG \) is congruent to chord \( EF \) (marked with same tick), the central angle for arc \( FG \) (which is \( \angle FCG \)) is equal to \( \angle ECF \), so \( m\overarc{FG} = 67^\circ \), and \( m\angle ECF = 67^\circ \). Wait, but let's confirm. The sum of central angles: if we have \( \angle DCE = 75^\circ \) (given), \( \angle ECF = 67^\circ \), \( \angle FCG = 67^\circ \), then total would be \( 75 + 67 + 67 = 209^\circ \), but maybe there's another angle. Wait, no, the diagram shows \( DE \cong EF \cong FG \), so their central angles are equal. So \( \angle DCE = \angle ECF = \angle FCG \)? Wait, no, \( DE \) is congruent to \( EF \), and \( EF \) is congruent to \( FG \), so all three chords are congruent, so their central angles are equal. Wait, but the given angle \( 75^\circ \) – maybe \( \angle DCE = 75^\circ \), then \( \angle ECF = 67^\circ \), \( \angle FCG = 67^\circ \), and the remaining angle would be \( 360 - 75 - 67 - 67 = 151^\circ \), but that's not necessary. The key is that congruent chords have congruent central angles, so arc \( FG \) has the same measure as arc \( EF \), which is \( 67^\circ \), and \( \angle ECF \) is the central angle for arc \( EF \), so it's \( 67^\circ \).

Step2: Confirm \( m\overarc{FG} \) and \( m\angle ECF \)

Since chords \( EF \) and \( FG \) are congruent, their subtended arcs \( \overarc{EF} \) and \( \overarc{FG} \) are congruent. The central angle \( \angle ECF \) subtends arc \( EF \), and we found \( \angle ECF = 67^\circ \) (from the isoceles triangle \( ECF \) with base angles \( 56.5^\circ \)). Thus, \( m\overarc{FG} = m\overarc{EF} = 67^\circ \), and \( m\angle ECF = 67^\circ \).

Answer:

\( m\overarc{FG} = \boxed{67^\circ} \)
\( m\angle ECF = \boxed{67^\circ} \)