WAEC - Mathematics (1999 - No. 37)

In the diagram, DE||BC, |AD| = x cm and |DB| = |AE| = ycm. Find |CE| in terms of x and y

\(\frac{x^2}{y}\)

\(\frac{y}{x^2}\)

\(\frac{y^2}{x}\)

\(\frac{x}{x+y}=\frac{y}{x+CE}\\
xy + xCE = xy + y^2\\
∴xCE = y^2\\
CE=\frac{y^2}{x}\)