ABCD is a rectangle in which BC = 2AB a point E lies on ray CD such that CE - Maths - Triangles

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

Shridhar Kaushik · Accepted Answer

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°

$and \frac{AB}{BC} = \frac{BC}{CE} = \frac{1}{2}$

∴ Triangles ABC and BCE are similar.

∴ ∠FAB = ∠FBC ..... (1)

and ∠BEC = ∠ACB ..... (2)

⇒ ∠ACB = ∠BEC ..... (3)

Adding (1) and (3), we get –

$∠ FAB + ∠ ACB = ∠ FBC + ∠ BEC \Rightarrow 90 ° = ∠ FBC + ∠ BEC \Rightarrow ∠ FBC + ∠ BEC = 90 ° \Rightarrow ∠ FAB + ∠ BEC = 90 ° [Using (1)] \Rightarrow ∠ FCE + ∠ BEC = 90 ° [∵ AB||CD]$

Using angle sum property of triangle we get –

∠EFC = 90°

Hence BE ⊥ AC

37

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.