0
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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.

enter image description here

$\endgroup$
  • $\begingroup$ How do we know that triangle ABE and BCD are similar? $\endgroup$ – Clancy Mar 28 '14 at 13:30
  • $\begingroup$ 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. $\endgroup$ – Alijah Ahmed Mar 28 '14 at 13:45
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.