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

#1**+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

#1**+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 :

CPhill Aug 30, 2017