Let \(\triangle ABC\) be a right triangle with right angle at \(A\) and \(AC\) the shorter leg. Point \(D\) lies inside the triangle such that \(\angle ADC=90^\circ\) and \(\angle DCA=\angle ABC\). Extend \(BD\) to a point \(E\) so that \(\angle BEA=90^\circ\) and \(\angle DAE=\angle ABC\). Find the ratio \(BD:DE\).
Solution
Answer: \(3:1\)
Use angle chasing, the angle bisector theorem, and the Pythagorean theorem. The solution PDF gives this hint and the answer \(BD:DE=3:1\).