1. If \(\frac{a^2}{b + c} = \frac{b^2}{c + a} = \frac{c^2}{a + b} = 1\), then show that \(\frac{a}{a + 1} + \frac{b}{b + 1} + \frac{c}{c + 1} = 2\).

2. \[ \frac{a^2}{b + c} = \frac{b^2}{c + a} = \frac{c^2}{a + b} = 1 \] **Show that:** \[ \frac{1}{1 + a} = \frac{1}{1 + b} = \frac{1}{1 + c} = 1 \]

3. If \(\cfrac{a^2}{b+c} = \cfrac{b^2}{c+a} + \cfrac{c^2}{a+b} = 1\), then prove that \[ \cfrac{1}{1+a} + \cfrac{1}{1+b} + \cfrac{1}{1+c} = 1 \]

4. If \(\cfrac{a^2}{b+c} = \cfrac{b^2}{c+a} = \cfrac{c^2}{a+b} = 1\), then prove that \(\cfrac{1}{a+1} + \cfrac{1}{b+1} + \cfrac{1}{c+1} = 1\).

5. If \(\cfrac{a}{1-a}+\cfrac{b}{1-b}+\cfrac{c}{1-c}=1\), then prove that \(\cfrac{1}{1-a}+\cfrac{1}{1-b}+\cfrac{c}{1-c}=4\).

6. If \(a + \frac{1}{b} = 1\) and \(b + \frac{1}{c} = 1\), then prove that \(c + \frac{1}{a} = 1\).

7. If \[ \frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b} \] then prove that each of these ratios is either \(\frac{1}{2}\) or \(-1\).

8. If \(\cfrac{a + b − c}{a + b} = \cfrac{b + c − a}{b + c} = \cfrac{c + a − b}{c + a}\) and \(a + b + c ≠ 0\), then prove that \(a = b = c\).

9. 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 \[ \frac{a}{x} = \frac{b}{y} = \frac{c}{z} \]

10. 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 \[ \frac{x}{a} = \frac{y}{b} = \frac{z}{c} \]

11. If \(\cfrac{a}{b + c} = \cfrac{b}{c + a} = \cfrac{c}{a + b}\) and \(a + b + c \ne 0\), then prove that \(a = b = c\).

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

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

14. If \(a + b + c = 0\), then prove that \[ \frac{bc}{2a^2 + bc} + \frac{ca}{2b^2 + ca} + \frac{ab}{2c^2 + ab} = 1 \]

15. If \(a + \frac{1}{b} = 1\) and \(b + \frac{1}{c} = 1\), then show that \(c + \frac{1}{a} = 1\).

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

17. If the roots of the quadratic equation \((a^2 + b^2)x^2 + 2(ac + bd)x + (c^2 + d^2) = 0\) are equal, then prove that \(\cfrac{a}{b} = \cfrac{c}{d}\).

18. If \[ \cfrac{a+b}{b+c}=\cfrac{c+d}{d+a} \] then prove that \[ c=a \quad \text{or} \quad a+b+c+d=0. \]

19. If \(\cfrac{bz+cy}{a} = \cfrac{cx+az}{b} = \cfrac{ay+bx}{c}\), then prove that \(\cfrac{x}{a(b^2+c^2-a^2)} = \cfrac{y}{b(c^2+a^2-b^2)} = \cfrac{z}{c(a^2+b^2-c^2)}\).

20. If \(a + b + c = 0\), then prove that: \[ \cfrac{1}{2a^2 + bc} + \cfrac{1}{2b^2 + ca} + \cfrac{1}{2c^2 + ab} = 0 \]

21. If \(a + c = 2b\) and \(q^2 = pr\), then prove that \(p^{b-c} \cdot q^{c-a} \cdot r^{a-b} = 1\).

22. If \[ \cfrac{a^2}{b+c} = \cfrac{b^2}{c+a} = \cfrac{c^2}{a+b} = 1 \] then prove that \[ \cfrac{1}{a+1} + \cfrac{1}{b+1} + \cfrac{1}{c+1} = 1. \]

23. If \(\cfrac{a^2}{b+c} = \cfrac{b^2}{c+a} + \cfrac{c^2}{a+b} = 1\), prove that \(\cfrac{1}{1+a} + \cfrac{1}{1+b} + \cfrac{1}{1+c} = 1\).

24. If \(\cfrac{a^2}{b+c} = \cfrac{b^2}{c+a} + \cfrac{c^2}{a+b} = 1\), then show that \(\cfrac{1}{1+a} + \cfrac{1}{1+b} + \cfrac{1}{1+c} = 1\).

25. If \(\cfrac{a^2}{b+c} = \cfrac{b^2}{c+a} + \cfrac{c^2}{a+b} = 1\), then show that \[ \cfrac{1}{1+a} + \cfrac{1}{1+b} + \cfrac{1}{1+c} = 1 \]

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

27. If \(\cfrac{a}{3} = \cfrac{b}{4} = \cfrac{c}{7}\), then prove that \[ \cfrac{a + b + c}{c} = 2 \]

28. If \[ \cfrac{a + b}{b + c} = \cfrac{c + d}{d + a} \] then prove that either \(c = a\) or \(a + b + c + d = 0\)

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

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