Question

ngle below, c is the circumcen 2 f 7x - 29 31 e g 3x + 7 c 14 d a b 74

Explanation:

Step1: Recall Circumcenter Property

The circumcenter \( C \) of a triangle is equidistant from all vertices, so \( CD = CF = CB \). Also, the perpendicular from \( C \) to a side bisects that side. Here, \( CA \) is perpendicular to \( DB \), so \( DA = AB \). But we need to find \( x \) using \( DE = DG \)? Wait, no, looking at the segments: \( DE = 3x + 7 \) and \( EF = 7x - 29 \), but since \( C \) is circumcenter, \( CE \) bisects \( DF \)? Wait, actually, in a triangle, the circumcenter is the intersection of perpendicular bisectors. So \( CE \), \( CA \), \( CG \) are perpendicular bisectors? Wait, \( DA = AB = \frac{74}{2}=37 \), but maybe the key is that \( DE = EF \)? Wait, no, the segments from \( D \) to \( E \) is \( 3x + 7 \), and from \( E \) to \( F \) is \( 7x - 29 \), but since \( C \) is circumcenter, \( CE \) is a perpendicular bisector of \( DF \), so \( DE = EF \). Wait, no, \( DE \) and \( EF \): if \( CE \) is the perpendicular bisector, then \( DE = EF \). So set \( 3x + 7 = 7x - 29 \).

Step2: Solve for \( x \)

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Wait, but let's check. If \( x = 9 \), then \( 3x + 7 = 3*9 +7 = 34 \), \( 7x -29 = 63 -29 = 34 \). So that works. Alternatively, maybe the segments from \( D \) to \( C \) and \( F \) to \( C \) are equal? Wait, \( DC \): let's see, \( DA = 37 \), \( CA =14 \), so \( DC = \sqrt{37^2 +14^2} \), but maybe not. The key is the perpendicular bisector: \( CE \) bisects \( DF \), so \( DE = EF \), hence \( 3x +7 =7x -29 \), solving gives \( x=9 \).

Answer:

\( x = 9 \)