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.
Given : ABCD is a rectangle in which BC = 2 AB.
CE = 2BC
To Prove : BE ⊥ AC
Let AB = x
Then, BC = 2x
and CE = 2(BC) = 4x.
Also ∠B = ∠C = 90°
∴ Triangles ABC and BCE are similar.
∴ ∠FAB = ∠FBC ..... (1)
and ∠BEC = ∠ACB ..... (2)
⇒ ∠ACB = ∠BEC ..... (3)
Adding (1) and (3), we get –
Using angle sum property of triangle we get –
∠EFC = 90°
Hence BE ⊥ AC