In figure 1.81, the vertices of square DEFG are on the sides of \(\triangle ABC\). \(\angle A = 90^\circ\). Then prove that \(DE^2 = BD \times EC\).
(Hint: Show that \(\triangle GBD\) is similar to \(\triangle CFE\). Use GD = FE = DE.)
Figure: D and E lie on side BC (B–D–E–C), G lies on side AB and F lies on side AC.
Answer: \(\triangle GBD \sim \triangle CFE\) (AA test), so \(\frac{GD}{CE} = \frac{BD}{FE}\); with GD = FE = DE this gives \(DE^2 = BD \times EC\).
DEFG is a square, so \(\angle GDB = \angle FEC = 90^\circ\) and GD = FE = DE.
In \(\triangle ABC\), \(\angle A = 90^\circ\), so \(\angle B + \angle C = 90^\circ\).
In \(\triangle FEC\), \(\angle FEC = 90^\circ\), so \(\angle CFE + \angle C = 90^\circ\). Hence \(\angle CFE = \angle B\), i.e. \(\angle GBD \cong \angle CFE\).
In \(\triangle GBD\) and \(\triangle CFE\):
\(\therefore \triangle GBD \sim \triangle CFE\) (AA test), with G ↔ C, B ↔ F, D ↔ E.
\[\frac{GD}{CE} = \frac{BD}{FE}\] \[GD \times FE = BD \times CE\]Since GD = FE = DE, \(DE^2 = BD \times EC\). Hence proved.
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