Triangle ABC has a right angle at angle 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.

Below is the diagram representing the described triangle.

Can you please provide a step-by-step solution?

Also, it would be helpful if you could solve this without trignometry, because that is too complex for someone with my understanding of geometry. I think that you prove similarity by SSS, SAS, or AA, but I am not sure.

Thank you for your help!

Guest Mar 10, 2017

#1**+5 **

**Triangle ABC has a right angle at angle 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.**

**Below is the diagram representing the described triangle.**

**Geometric mean theorem:**

\(\begin{array}{|lrcll|} \hline (1) & AB^2 &=& EB \cdot BC \\ (2) & BC^2 &=& AB\cdot BD \\ \hline (1) \cdot (2): & AB^2\cdot BC^2 &=& EB \cdot BC\cdot AB\cdot BD \quad & | \quad : (AB\cdot BC) \\ & AB\cdot BC &=& EB\cdot BD \\ \hline \end{array}\)

heureka
Mar 10, 2017

#1**+5 **

Best Answer

**Triangle ABC has a right angle at angle 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.**

**Below is the diagram representing the described triangle.**

**Geometric mean theorem:**

\(\begin{array}{|lrcll|} \hline (1) & AB^2 &=& EB \cdot BC \\ (2) & BC^2 &=& AB\cdot BD \\ \hline (1) \cdot (2): & AB^2\cdot BC^2 &=& EB \cdot BC\cdot AB\cdot BD \quad & | \quad : (AB\cdot BC) \\ & AB\cdot BC &=& EB\cdot BD \\ \hline \end{array}\)

heureka
Mar 10, 2017