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# Geometry question

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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 $$\angle EAB \cong \angle ACB$$.

Oct 24, 2018

#1
+539
+11

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
#2
+539
+12

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.

Oct 25, 2018
#3
+94321
+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

Oct 25, 2018