Question

ABCD is a rectangle in which BC = 2AB. a point E lies on ray CD such that CE = 2BC. prove that BE is perpendicular to AC.

Answer 1

See the diagram.
We will prove by using similar triangles and Pythagoras theorem.

Let AB = 1 unit.        So, BC  = 2  and   AC = rt5
       and also, CE = 2 * BC = 4

Using Pythagoras theorem in triangle BCE:   BE = √(2²+4²)=2 √5

Triangles AOB & COE are similar, as AO || CO, OB || OE and AB || CE.
    Hence,    CE= 4 AB,    CO = 4 AO, and  OE = 4 BO 

Thus,  CO = 4/5  * AC =  4/√5   
          OE = 4/5 * BE = 4/5 * 2√5 = 8 /√5

In triangle CEO:   CO² + OE² =16/5  + 64/5 = 16  = CE².
Hence, as per Pythagorean theorem,  Angle COE = 90 deg.

Answer 2

Answer:

To prove BE is perpendicular to AC, it is enough to show <COE=90°

First prove ∆ABC~∆BCE by SAS

Then prove ∆ABC~∆COC by AA