Geometry/Probability Problem

Triangle ABC is 30-60-90 

AB = 2        BC = AB/2 = 1     AC =sqrt [(AB)² - (BC)²] = 1.732050808

CD = sqrt[(BD)² - (BC)²] = 1

Probability that BD < sqrt(2) = (CD) / (AC) = 0.577350269  or  57.7350269 %

I've corrected my error with  red "ink"   

Nobody's perfect 

The restriction $BD<\sqrt{2}$ creates a 45 - 45 - 90 triangle WITHIN the 30 - 60 - 90 triangle.

We easily calculate the area of the 30 - 60 - 90 triangle.

Small leg = 1

Long leg = $\sqrt{3}$

Area = $\frac{\sqrt{3}}{2}$

We easily calculate the area of the 45 - 45 - 90 triangle.

Leg = 1

Leg = 1

Area = $\frac{1}{2}$

Now the probability of BD being less than root 2 is the area of the 45 - 45 - 90 triangle divided by 30 - 60 - 90 triangle.

Now you do it!

Yes please don't just give an answer to me, If you are, at least give an explanation too!

A  =  30°

D                2

1

C          1                 B  =  60°

BC  =  1

AC  =   √3

If   BD  = √2

Then  DC  =   √[ (2)  - 1 ] =  1

The  area of  triangle  ABC  = (1/2) BC)(AC)  =  (1/2)(1)(√3) =  √3/2

Triangle  DCB  will  have  a max  area  when BD  = √2

And  its area  =  (1/2)(CD)(CB)  = (1/2)(1)(1)  = 1/2

So......the probability  that  BD < √2   will occur  when P  falls inside triangle  DCB

And this probability is     (1/2)  / (√3) /2  =   1 / √3