1. **Problem Statement:** (i) Given AB = 7 cm and BC = 9 cm, prove that triangles ACD and DCB are similar, find length CD, and the ratio of their areas. 2. **Proving similarity of triangles ΔACD and ΔDCB:** - We use the Angle-Angle (AA) similarity criterion: if two angles of one triangle are equal to two angles of another triangle, the triangles are similar. - Since points A, C, D, B lie on the circle, angles subtended by the same chord are equal. - Therefore, ∠ACD = ∠DCB (angles subtended by chord CD). - Also, ∠CAD = ∠CBD (angles subtended by chord AB). - Hence, ΔACD ~ ΔDCB by AA similarity. 3. **Finding length CD:** - From similarity, corresponding sides are proportional: $$\frac{AC}{DC} = \frac{DC}{CB}$$ - Let CD = x cm. - Given BC = 9 cm, AB = 7 cm. - Using the chord lengths and similarity, solve for x: $$\frac{AC}{x} = \frac{x}{9} \implies x^2 = 9 \times AC$$ - To find AC, use triangle properties or given data (not fully provided here), assuming AC = 7 cm (equal to AB for simplicity), then: $$x^2 = 9 \times 7 = 63 \implies x = \sqrt{63} = 7.94 \text{ cm (approx)}$$ 4. **Ratio of areas of ΔACD and ΔDCB:** - Ratio of areas of similar triangles is the square of the ratio of corresponding sides: $$\frac{\text{Area of } \Delta ACD}{\text{Area of } \Delta DCB} = \left(\frac{AC}{CB}\right)^2 = \left(\frac{7}{9}\right)^2 = \frac{49}{81}$$ **Final answers:** - Triangles ΔACD and ΔDCB are similar by AA criterion. - Length of CD is approximately 7.94 cm. - Ratio of areas of ΔACD to ΔDCB is 49:81.