Geometry question

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

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

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.

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