If b is the mean proportional between a and c, prove that a, c, \[{{\mathbf{a}}^{\mathbf{2}}}~+\text{ }{{\mathbf{b}}^{\mathbf{2}}}~\mathbf{and}\text{ }{{\mathbf{b}}^{\mathbf{2}}}~+\text{ }{{\mathbf{c}}^{\mathbf{2}}}~\] are proportional.

Solution:

It is given that

b is the mean proportional between a and c

We can write it as

b² = a × c

b² = ac ….. (1)

We know that

a, c, \[{{\mathbf{a}}^{\mathbf{2}}}~+\text{ }{{\mathbf{b}}^{\mathbf{2}}}~\mathbf{and}\text{ }{{\mathbf{b}}^{\mathbf{2}}}~+\text{ }{{\mathbf{c}}^{\mathbf{2}}}~\]are in proportion

It can be written as

\[a/c\text{ }=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}})/\text{ }({{b}^{2}}~+\text{ }{{c}^{2}})\]

By cross multiplication

\[a\text{ }({{b}^{2}}~+\text{ }{{c}^{2}})\text{ }=\text{ }c\text{ }({{a}^{2}}~+\text{ }{{b}^{2}})\]

Using equation (1)

\[a\text{ }(ac\text{ }+\text{ }{{c}^{2}})\text{ }=\text{ }c\text{ }({{a}^{2}}~+\text{ }ac)\]

So we get

\[ac\text{ }\left( a\text{ }+\text{ }c \right)\text{ }=\text{ }{{a}^{2}}c\text{ }+\text{ }a{{c}^{2}}\]

Here ac (a + c) = ac (a + c) which is true.

Therefore, it is proved.