1. **Problem Statement:** Prove the given geometric properties and find lengths in the circle with tangent AC at C, chord DF extended to A, points B and G on AC and AD such that GB || CF, and intersection E of CF and DB. 2. **Step 3.1.1: Prove $\overset{\frown}{C_1} = \overset{\frown}{BGD}$** - Since AC is tangent at C, angle between tangent AC and chord CF equals the angle in the alternate segment, so $\angle ACF = \angle C_1$. - Given GB || CF, alternate interior angles $\angle BGD = \angle C_1$. - Therefore, arcs subtended by these equal angles are equal: $\overset{\frown}{C_1} = \overset{\frown}{BGD}$. 3. **Step 3.1.2: Prove BCDG is cyclic** - Since $\overset{\frown}{C_1} = \overset{\frown}{BGD}$, angles subtended by these arcs are equal. - Opposite angles in quadrilateral BCDG sum to 180°, so BCDG is cyclic. 4. **Step 3.2: Prove $\triangle DDCB ||| ACFG$** - Given GB || CF and B, G lie on AC and AD respectively. - By corresponding angles and parallel lines, $\triangle DDCB$ and $\triangle ACFG$ have equal angles. - Using similarity criteria (AA), $\triangle DDCB \sim \triangle ACFG$. 5. **Step 3.3: Find length of CF given $CD=12$ and $3BC=4FG$** - From similarity, ratios of sides hold: $\frac{BC}{FG} = \frac{CD}{CF}$. - Given $3BC=4FG \Rightarrow \frac{BC}{FG} = \frac{4}{3}$. - Substitute: $\frac{4}{3} = \frac{12}{CF} \Rightarrow CF = \frac{12 \times 3}{4} = 9$ units. 6. **Step 3.4: Find length of DE given $BD=16$, $DF=13$, $BC=8$** - Since BCDG is cyclic and E lies on DB and CF, use power of point or intersecting chords theorem: $$ BD \times DE = CF \times FE $$ - From previous, $CF=9$ units. - Also, $DF = DE + EF = 13$. - Let $DE = x$, then $EF = 13 - x$. - Substitute: $$ 16 \times x = 9 \times (13 - x) $$ $$ 16x = 117 - 9x $$ $$ 16x + 9x = 117 $$ $$ 25x = 117 $$ $$ x = \frac{117}{25} = 4.68 $$ units. **Final answers:** - $\overset{\frown}{C_1} = \overset{\frown}{BGD}$ - BCDG is cyclic - $\triangle DDCB \sim \triangle ACFG$ - $CF = 9$ units - $DE = 4.68$ units