Question

If by+cz/b²+c²=cz+ax/c²+a²=ax+by/a²+b² then prove that x/a=y/b=z/c

Answer 1

Answer:

by+cz/b²+c²=cz+ax/c²+a²=ax+by/a²+b²  = K  = x/a=y/b=z/c

Step-by-step explanation:

let say x/a=y/b=z/c = K

Then x = ak

y = bk

& z = ck

by+cz/b²+c²= b²k + c²k /b²+c²  = k(b²+c²)/(b²+c²) = k

cz+ax/c²+a²= c²k + a²k /c²+a²  = k(c²+a²)/(c²+a²) = k

ax+by/a²+b² = a²k + b²k /a²+b²  = k(a²+b²)/(a²+b²) = k

=> by+cz/b²+c²=cz+ax/c²+a²=ax+by/a²+b²  = K  = x/a=y/b=z/c

Answer 2

Answer:

cz+ax/c²+a²=ax+by/a²+b² then prove that x/a=y/b=z/c