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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!

Guest Oct 24, 2018
 #1
avatar+406 
+7

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

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

KnockOut  Oct 25, 2018
 #2
avatar+406 
+8

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.

KnockOut  Oct 25, 2018
 #3
avatar+90995 
+1

Thanks, Knockout...!!!!

 

Look at the figure below

 

 

 ∠ EBA  =  ∠ DBC  =   90°

 ∠ EAB  = ∠ CDB 

 

So....by  AA congruency  

 

Δ BAE   ~  ΔBDC  

 

So

 

AB / BE  =  BD / BC  

 

AB * BC   =  BD * BE

 

EB * BD    =  AB * BC

 

 

 

cool cool cool

CPhill  Oct 25, 2018

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