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# Similar Triangles

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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 $$EB\cdot BD=AB\cdot BC$$.

Aug 30, 2017

#1
+100516
+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 :

Aug 30, 2017

#1
+100516
+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 :

CPhill Aug 30, 2017