+436 In rectangle $ABCD$, points $E$ and $F$ lie on segments $AB$ and $CD$, respectively, such that $AE = \frac{AB}{8}$ and $CF = \frac{CD}{5}$. Segment $BD$ intersects segment $EF$ at $P$. What fraction of the area of rectangle $ABCD$ lies in triangle $EBP$? Express your answer as a common fraction.
+129771 
Let D= (0,0) Let B= (40,10)
Slope of line through DB = ( 10 - 0 ) / (40 -0) = 1/4
Equation of this line
y =(1/4)x
Let E = (5,10) Let F = (8,0)
Slope of line through EF = (0 -10)/ (8-5) = -10/3
Equation of this line
y = (-103)(x -8)
y = (-10/3)x + 80/3
To find the x coordinate of P
(1/4)x = (-10/3)x + 80/3
(1/4 + 10/3)X = 80/3
(3/12 + 40/12)x = 80/3
(43/12)x = 80/3
x = (80/3) (12/43) = 320/43
y coordinate of P = (1/4) (320/43) = 80/43
Height of triangle EBP = 10 - 80/43 = 350/43
Base of EBP = 40 - 5 = 35
Area of EBP = (1/2) (35) (350/43) = 6125/43
[EBP ] / [ ABCD] = (6125/43) / [400] = 245 / 688
