1. If \[ \cfrac{x^2 - yz}{a} = \cfrac{y^2 - zx}{b} = \cfrac{z^2 - xy}{c}, \] then prove that \[ (a + b + c)(x + y + z) = ax + by + cz. \]

2. If \[ \cfrac{x^2 - yz}{a} = \cfrac{y^2 - zx}{b} = \cfrac{z^2 - xy}{c}, \] then prove that \[ (a + b + c)(x + y + z) = ax + by + cz. \]

3. If \[ \frac{x^2}{by + cz} = \frac{y^2}{cz + ax} = \frac{z^2}{ax + by} = 1 \] then prove that \[ \frac{a}{a + x} + \frac{b}{b + y} + \frac{c}{c + z} = 1 \]

4. If \(\cfrac{x^2 - yz}{a} = \cfrac{y^2 - zx}{b} = \cfrac{z^2 - xy}{c}\), then prove that > \((a + b + c)(x + y + z) = ax + by + cz\)

5. If \(a^2 = by + cz\), \(b^2 = cz + ax\), and \(c^2 = ax + by\), then prove that \[ \frac{x}{a + x} + \frac{y}{b + y} + \frac{z}{c + z} = 1 \]

6. If \[ x^2 : (by + cz) = y^2 : (cz + ax) = z^2 : (ax + by) = 1 \] then show that \[ \frac{a}{a + x} + \frac{b}{b + y} + \frac{c}{c + z} = 1 \]

7. If \(bcx = cay = abz\), then prove that \[ \frac{ax + by}{a^2 + b^2} = \frac{by + cz}{b^2 + c^2} = \frac{cz + ax}{c^2 + a^2} \]

8. If \[ \frac{by + cz}{b^2 + c^2} = \frac{cz + ax}{c^2 + a^2} = \frac{ax + by}{a^2 + b^2} \] then prove that \[ \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \]

9. Given: \[ x^2 : (by + cz) = y^2 : (cz + ax) = z^2 : (ax + by) = 1 \] Show that: \[ \frac{a}{a + x} + \frac{b}{b + y} + \frac{c}{c + z} = 1 \]

10. If \(x = cy + bz\), \(y = az + cx\), and \(z = bx + ay\), then prove that \[ \frac{x^2}{1 - a^2} = \frac{y^2}{1 - b^2} = \frac{z^2}{1 - c^2} \]

11. If \[ \cfrac{x + y}{3a - b} = \cfrac{y + z}{3b - c} = \cfrac{z + x}{3c - a} \] then prove that \[ \cfrac{x + y + z}{a + b + c} = \cfrac{ax + by + cz}{a^2 + b^2 + c^2} \]

12. If \(x = cy + bz\), \(y = az + cx\), and \(z = bx + ay\), then prove that \[ \frac{x^2}{1 - a^2} = \frac{y^2}{1 - b^2} \]

13. If \[ \cfrac{ax + by}{a} = \cfrac{bx - ay}{b} \] then prove that each ratio is equal to \(x\).

14. If \[ \cfrac{x^2}{by + cz} = \cfrac{y^2}{cz + ax} = \cfrac{z^2}{ax + by} = 1 \] then show that \[ \cfrac{a}{a + x} + \cfrac{b}{b + y} + \cfrac{c}{c + z} = 1 \]

15. If \[ \frac{a}{y + z} = \frac{b}{z + x} = \frac{c}{x + y} \] then prove that \[ \frac{a(b - c)}{y^2 - z^2} = \frac{b(c - a)}{z^2 - x^2} = \frac{c(a - b)}{x^2 - y^2} \]

16. If the ratio of the roots of the quadratic equation \(ax^2 + bx + c = 0\) is \(1 : r\), then prove that \[ \frac{(r + 1)^2}{r} = \frac{b^2}{ac} \]

17. If \(x : a = y : b = z : c\), then prove that \((a^2 + b^2 + c^2)(x^2 + y^2 + z^2) = (ax + by + cz)^2\)

18. If \(\cfrac{a}{b} = \cfrac{x}{y}\), then prove that \[ (a + b)(a^2 + b^2)x^3 = (x + y)(x^2 + y^2)a^3 \]

19. If \[ \cfrac{x}{b + c - a} = \cfrac{y}{c + a - b} = \cfrac{z}{a + b - c} \] then prove that \[ (b - c)x + (c - a)y + (a - b)z = 0 \]

20. If \[ \cfrac{a}{y + z} = \cfrac{b}{z + x} = \cfrac{c}{x + y}, \] then prove that \[ \cfrac{a(b - c)}{y^2 - z^2} = \cfrac{b(c - a)}{z^2 - x^2} = \cfrac{c(a - b)}{x^2 - y^2}. \]

21. If \(x = cy + bz\), \(y = az + cx\), and \(z = bx + ay\), then show that: \[ \frac{x^2}{1 - a^2} = \frac{y^2}{1 - b^2} \]

22. If \[ (b + c - a) x = (c + a - b) y = (a + b - c) z = 2 \] then prove that \[ \left( \frac{1}{y} + \frac{1}{z} \right) \left( \frac{1}{z} + \frac{1}{x} \right) \left( \frac{1}{x} + \frac{1}{y} \right) = abc \]

23. If \(\frac{x}{y} = \frac{a + 2}{a - 2}\), then prove that \[ \frac{x^2 - y^2}{x^2 + y^2} = \frac{4a}{a^2 + 4} \]

24. If \( \cot\theta = \frac{x}{y} \), then prove that \[ \frac{x\cos\theta - y\sin\theta}{x\cos\theta + y\sin\theta} = \frac{x^2 - y^2}{x^2 + y^2} \]

25. If \((b + c - a)x = (c + a - b)y = (a + b - c)z = 2\), then show that \[ \left(\frac{1}{x} + \frac{1}{y}\right)\left(\frac{1}{y} + \frac{1}{z}\right)\left(\frac{1}{z} + \frac{1}{x}\right) = abc \]

26. If \(x : a = y : b = z : c\), then show that \((a^2 + b^2 + c^2)(x^2 + y^2 + z^2) = (ax + by + cz)^2\)

27. If \[ \frac{b + c - a}{y + z - x} = \frac{c + a - b}{z + x - y} = \frac{a + b - c}{x + y - z} \] then prove that \[ \frac{a}{x} = \frac{b}{y} = \frac{c}{z} \]

28. If ABCD is a cyclic quadrilateral inscribed in a circle with center O, then prove that: \[ AB + CD = BC + DA \]

29. If \(a : b = b : c\), then prove that \[ (a + b)^2 : (b + c)^2 = a : c \]

30. If \(a, b, c\) are in continued geometric progression (i.e., in successive proportion), then prove that \[ \cfrac{1}{b^2} = \cfrac{1}{b^2 - a^2} + \cfrac{1}{b^2 - c^2} \]