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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!

 Oct 24, 2018

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

\(\large \frac{EB}{AB}=\frac{BC}{BD}\Rightarrow EB\cdot BD=AB\cdot BC\).

 Oct 25, 2018

The figure below can be helpful in understanding the solution.




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.

 Oct 25, 2018

Thanks, Knockout...!!!!


Look at the figure below



 ∠ EBA  =  ∠ DBC  =   90°

 ∠ EAB  = ∠ CDB 


So....by  AA congruency  


Δ BAE   ~  ΔBDC  




AB / BE  =  BD / BC  


AB * BC   =  BD * BE


EB * BD    =  AB * BC




cool cool cool

 Oct 25, 2018

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