Triangle ABC has a right angle at B. Legs AB and CB are extended past point B to points D and E, respectively, such that angle EAC = angle ACD = 90 degrees. Prove that EB * BD = AB * BC.
E B C
Angle AEB = Angle BAC
Angle ABE = Angle ABC
Thus, by AA congruency, triangle EBA is similar to triangle ABC
Angle CAB = Angle BCD
Angle CBA = Angle CBD
Again, by AA congruency, triangle ABC is similar to triangle CBD
So..by transivity......triangle EBA is similar to triangle CBD
EB / BA = CB / BD
EB * BD = BA * CB which is the same as.....
EB * BD = AB * BC