Prove that a:b=c:da:b = c:da:b=c:d , if
4a+5b4a−5b=4c+5d4c−5d(ii)2a+9b2a−9b=2c+9d2c−9d4a+5b4a−5b=4c+5d4c−5d(ii)2a+9b2a−9b=2c+9d2c−9d\frac{4a + 5b}{4a - 5b} = \frac{4c + 5d}{4c - 5d} \qquad (ii) \frac{2a + 9b}{2a - 9b} = \frac{2c + 9d}{2c - 9d}4a−5b4a+5b=4c−5d4c+5d(ii)2a−9b2a+9b=2c−9d2c+9d4a−5b4a+5b=4c−5d4c+5d(ii)2a−9b2a+9b=2c−9d2c+9d
ac2+bd2ac2−bd2=c3+d3c3−d3(iv)a2c+b2da2c−b2d=ac2+bd2ac2−bd2ac2+bd2ac2−bd2=c3+d3c3−d3(iv)a2c+b2da2c−b2d=ac2+bd2ac2−bd2\frac{ac^2 + bd^2}{ac^2 - bd^2} = \frac{c^3 + d^3}{c^3 - d^3} \qquad (iv) \frac{a^2c + b^2d}{a^2c - b^2d} = \frac{ac^2 + bd^2}{ac^2 - bd^2}ac2−bd2ac2+bd2=c3−d3c3+d3(iv)a2c−b2da2c+b2d=ac2−bd2ac2+bd2ac2−bd2ac2+bd2=c3−d3c3+d3(iv)a2c−b2da2c+b2d=ac2−bd2ac2+bd2
pa+qb:pa−qb=pc+qd:pc−qdpa + qb:pa - qb = pc + qd:pc - qdpa+qb:pa−qb=pc+qd:pc−qd
a+b+c+da+b−c−d=a−b+c−da−b−c+da+b+c+da+b−c−d=a−b+c−da−b−c+d\frac{a + b + c + d}{a + b - c - d} = \frac{a - b + c - d}{a - b - c + d}a+b−c−da+b+c+d=a−b−c+da−b+c−da+b−c−da+b+c+d=a−b−c+da−b+c−d