Triangle \(ABC\) has a right angle at \(B\). Legs \(\overline {AB}\) and \(\overline {CB}\) are extended past point \(B\) to points \(D\) and \(E\), respectively, such that \(\angle EAC = \angle ACD = 90^\circ\). Prove that \(EB \cdot BD = AB \cdot BC\)
I understand that I need to use similar triangles for this one, which I am trying with \(\triangle EBA \) and \(\triangle ABC\) with AA similarity. Now I am stuck on proving \(\angle EAB \cong \angle ACB\).
Thanks in advance!
+997 Consider similar triangles \(ABE\) and \(BCD\), then we have \(\angle BDC=\angle EAB\).
For triangle \(ABE\),
\(\tan(\angle EAB)=\frac{EB}{AB}\).
For triangle \(BCD\),
\(\tan(\angle BDC)=\frac{BC}{BD}\)
Thus we have
\(\large \frac{EB}{AB}=\frac{BC}{BD}\Rightarrow EB\cdot BD=AB\cdot BC\).
+997 The figure below can be helpful in understanding the solution.
https://i.stack.imgur.com/2SopB.png
We are given \(\angle ACD = \angle EAC= 90^{\circ}\), so \(AE\) is parallel to \(CD\), so \(\angle EAB = \angle BDC\).
We also have \(\angle EBA= \angle CBD = 90^{\circ}\), so that \(\angle AEB = \angle BCD\).
So angles of triangles \(ABE\) and \(BCE\) are the same, hence they are similar.
