Question 1210173
<pre>
Ikleyn's solution is incorrect.  But she refuses to admit her error.  
She apparently thinks there is a theorem that states:

{{{cross(matrix(4,1,
matrix(1,8,Any,line,drawn,from,the,vertex, to, the), 
matrix(1,6,opposite, side, of, a, triangle, divides),
matrix(1,7,the, opposite, side, into, the, ratio, of), 
matrix(1,7,the,areas,of,the,two,smaller,triangles)))}}}

There is no such theorem but Ikleyn believed she could
use it anyway to show that

{{{cross("AD'"/AC = matrix(1,4,Area, of, triangle,"ABD'")/matrix(1,4,Area, of, triangle,ABC))}}}

Below is the correct solution.  It's very long, which is why I did not have
time to finish the other day, and thought someone else might take over.  I
have checked it thoroughly and have found no errors. 

Ikleyn made a bad assumption, but when I challenged her, she wanted to fight,
and accused me of being the one wanting to fight.
{{{drawing(400,1600/9,-1,8,-1,3,

triangle(0,0,7,0,2.357142857,1.855768722),

green(line(0,0,3.78882,1.28251)),

locate(0,0,B),locate(7,0,C), locate(2.27,2.18,A),
green(locate(3.77,1.6,"D'")),

locate(1.2,1.53,3), locate(3.5,0,7), locate(4.5,1.4,5),

green(locate(2.2,.8,4)), red(line(0,0,2,1),
circle(2,1,.1), locate(2,1.4,P),
line(3.1098313,1.5549156,2.1,1.05),locate(3,2,D))
) 


 )}}}

We use the law of cosines to find angle A:

{{{BC^2=AB^2+AC^2-2*AB*AC*cos(A)}}}
{{{7^2=3^2+5^2-2*3*5*cos(A)}}}
{{{49=9+25-30*cos(A)}}}
{{{49=34-30*cos(A)}}}
{{{15=-30*cos(A)}}}
{{{15/(-30)=cos(A)}}}
{{{-1/2=cos(A)}}}
{{{A=120^o}}}

We find the area of triangle ABC

{{{Area = expr(1/2)*AB*AC*sin(A)}}}
{{{Area = expr(1/2)*3*5*sin(120^o)}}}
{{{Area = expr(15/2)*expr(sqrt(3)/2)}}}
{{{Area = expr(15/4)*sqrt(3)}}}

We use the law of cosines on triangle ABD' to find AD':
(note: We could use the law of sines, but the law of cosines
will give us AD' more directly)

{{{"BD'"^2=AB^2+"AD'"^2-2*AB*"AD'"*cos(A)}}}
{{{4^2=3^2+"AD'"^2-2*3*"AD'"*cos(120^o)}}}
{{{16=9+"AD'"^2-6*"AD'"*(-1/2)}}}
{{{7="AD'"^2+3*"AD'"}}}
{{{"AD'"^2+3*"AD'"-7=0}}}
{{{"AD'" = (-3 +- sqrt(3^2-4(1)(-7)))/(2*1)}}}
{{{"AD'" = (-3 +- sqrt(9+28))/2}}}
{{{"AD'" = (-3 +- sqrt(37))/2}}}
{{{"AD'" = expr(1/2)(-3 +- sqrt(37))}}}, ignore the minus sign:
{{{"AD'" = expr(1/2)(-3 + sqrt(37))}}}
{{{"AD'" = expr(1/2)(sqrt(37)-3)}}}
We find the area of triangle ABD'

{{{Area = expr(1/2)*AB*"AD'"*sin(A)}}}
{{{Area = expr(1/2)*3*(expr(1/2)(sqrt(37)-3))*sin(120^o)}}}
{{{Area = expr(1/2)*3*(expr(1/2)(sqrt(37)-3))*expr(sqrt(3)/2)}}}
{{{Area = expr((3sqrt(3))/8)*(sqrt(37)-3)}}}
{{{Area = expr(3/8)*(sqrt(3)*sqrt(37)-3*sqrt(3))}}}
{{{Area = expr(3/8)*(sqrt(111)-3*sqrt(3))}}}

The desired probability is the ratio of the area of triangle ABD' to
the area of triangle ABC.

{{{(expr(3/8)*(sqrt(111)-3*sqrt(3)))/(expr(15/4)*sqrt(3))}}}

Multiply top and bottom by 8

{{{(3(sqrt(111)-3*sqrt(3)))/(30sqrt(3))}}}
{{{(sqrt(111)-3*sqrt(3))/(10sqrt(3))}}}
{{{expr(1/10)*((sqrt(111)-3*sqrt(3))/(sqrt(3))))}}}
{{{expr(1/10)*(sqrt(111)/sqrt(3) - 3*sqrt(3)/sqrt(3))}}}
{{{expr(1/10)*(sqrt(111/3) - 3)}}}
{{{expr(1/10)*(sqrt(37) - 3)}}}  <-- correct answer, the probability that BD < 4.
That is approximately 0.308276253

Edwin</pre>