Triangle ABC has a right angle at B. Legs {AB} and {CB} are extended past point B to points D and E, respectively, such that < EAC = < ACD = 90 degrees. Prove that EB * BD = AB * BC.

I have tried to use similar triangles to solve problem (AA, SAS, SSS) but I can't seem to figure it out!

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.

• How do we know that triangle ABE and BCD are similar? – Clancy Mar 28 '14 at 13:30
• 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 $BCD$ are the same, hence they are similar. – Alijah Ahmed Mar 28 '14 at 13:45

Draw a figure. All triangles appearing therein are similar. It follows that $${|BE|\over |BA|}={|BC|\over|BD|}\ .$$