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Solution

Consider the following diagram.

As the two marked angles are corresponding we can deduce that the two triangles are similar.

$$\begin{eqnarray}\begin{eqnarray}\dfrac{a}{b} & = & \dfrac{x}{b-x}\\\therefore a(b-x) & = & bx\\ab - ax & = & bx\\ab & = & ab + bx\\& = & x(a + b)\\\therefore x & = & \dfrac{ab}{a+b}\end{eqnarray} && \textstyle{     OR     } &\begin{eqnarray}\dfrac{a-x}{x} & = & \dfrac{x}{b-x}\\\therefore (a-x)(b-x) & = & x^2\\ab - (a + b)x + x^2 & = & x^2\\(a + b)x & = & ab\\\therefore x & = & \dfrac{ab}{a+b}\end{eqnarray}\\\end{eqnarray}$$