+98 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 \(EB\cdot BD=AB\cdot BC\).
+128448
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 :
+128448
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 :
