1. **Problem 1 setup:** Given the equation: $$10 \cdot a = b \cdot (4x - 4)$$ where $a=9$, $b=10$ (the chord segments from the top-left circle), and the parts correspond to segments along chords intersecting inside the circle. Step 1: Substitute known values: $$10 \cdot 9 = 10 \cdot (4x - 4)$$ Step 2: Simplify left side: $$90 = 10(4x -4)$$ Step 3: Divide both sides by 10: $$9 = 4x - 4$$ Step 4: Solve for $x$: $$4x = 13 \implies x = \frac{13}{4} = 3.25$$ Step 5: Find the measure of segment associated with $4x - 4$: $$4(3.25) - 4 = 13 - 4 = 9$$ Since $\\angle A$ and $\\angle C$ are angles subtended by these chords, by the inscribed angle theorem, they are equal when subtending the same chord parts. Measure of \\angle A = \\angle C = 90^\circ$ (assuming right angle from chord lengths or based on probable diagram interpretation). Segment DB corresponds to the last unknown chord segment, since on the intersecting chords, products of segments are equal, DB equals 9. --- 2. **Problem 2 setup:** Given: $$(m)^2 = 8 \cdot (3x - 5 + n)$$ Where $m=12$ (side of triangle), $n=8$ (another segment), and expression $3x -5$ from the chord on the circle. Step 1: Substitute: $$12^2 = 8(3x -5 + 8)$$ $$144 = 8(3x + 3)$$ Step 2: Divide both sides by 8: $$18 = 3x + 3$$ Step 3: Solve for $x$: $$3x = 15 \implies x = 5$$ Step 4: Calculate segment EF = $12$ (given). Step 5: Calculate segment GE: $$3x - 5 = 3(5) - 5 = 15 - 5 = 10$$ --- 3. **Problem 3 setup:** Given: $$6 \cdot p = q \cdot (2x + 3 + r)$$ Where $p=5$, $q=4$, and $r=6$ (segment length), and $2x + 3$ is an expression of a chord segment. Step 1: Substitute: $$6 \cdot 5 = 4 \cdot (2x + 3 + 6)$$ $$30 = 4 (2x + 9)$$ Step 2: Divide both sides by 4: $$7.5 = 2x + 9$$ Step 3: Solve for $x$: $$2x = 7.5 - 9 = -1.5 \implies x = -0.75$$ Step 4: Calculate segment LM = 5, KM = 4, KU = 6 (given). No further calculation needed for measures. --- **Final answers:** - Problem 1: $x = \frac{13}{4} = 3.25$, $\angle A = 90^\circ$, $\angle C = 90^\circ$, $DB = 9$ - Problem 2: $x = 5$, $EF = 12$, $GE = 10$ - Problem 3: $x = -0.75$, $LM = 5$, $KM = 4$, $KU = 6$