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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 \(EB\cdot BD=AB\cdot BC\).

benjamingu22  Aug 30, 2017

Best Answer 

 #1
avatar+76038 
+1

 

Since ABC  is a right angle, then DBC is supplemental.....so it's = 90°as well

 

Snd EA is parallel to DC......and since transversal EC  cuts these, then angle DCE and angle CEA form equal alternate interior angles

 

Thus by angle - angle congruency......triangle CBD  is similar to triangle EBA

 

Thus BC / BD  =  BE / BA

 

So  BC * BA  =  BE * BD    or, put another way

 

AB * BC  =  EB * BD

 

Here's a pic :

 

 

 

cool cool cool

CPhill  Aug 30, 2017
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1+0 Answers

 #1
avatar+76038 
+1
Best Answer

 

Since ABC  is a right angle, then DBC is supplemental.....so it's = 90°as well

 

Snd EA is parallel to DC......and since transversal EC  cuts these, then angle DCE and angle CEA form equal alternate interior angles

 

Thus by angle - angle congruency......triangle CBD  is similar to triangle EBA

 

Thus BC / BD  =  BE / BA

 

So  BC * BA  =  BE * BD    or, put another way

 

AB * BC  =  EB * BD

 

Here's a pic :

 

 

 

cool cool cool

CPhill  Aug 30, 2017

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