Question

In the fig, ABCD is a rectangle, in which BC = 2AB. a point E lies oc CD produced such that CE= 2BC. find AC: BE.

Answer 1

We apply Pythagoras theorem to get the answer.

ΔABC is  a right angle triangle at B.
AC² = AB² + BC² = AB² + (2 AB)² = 5 AB²
AC = √5 AB

ΔBCE is a right angle triangle at C.
BE² = BC² + CE²  = BC² + (2 BC)² =  5  BC² = 5 (2 AB)² 
         = 20 AB²
BE = 2√5 AB

AC :  BE   = √5 AB : 2 √5 AB   =  1 : 2

Answer 2

In triangle ABC,

AC²=AB²+BC² (Pythagoras theorem)

AC²=AB²+(2AB)²

AC²=AB²+4AB²

AC²=5AB²

AC=√5 AB

In triangle BCE,

BE²=BC²+CE²

BE²=BC²+(2BC)²

BE²=BC²+4BC²

BE²=5BC²

BE =√5 BC

Then AC:BE=1:1