Help Please!!!!

I have down-scaled the problem a little, but the same logic would apply

The image below lays out the particulars :

The base of triangle XYZ  = XY  = 6  and the height  = 6...so...the area =

6^2 / 2 = 18 sq units

Now...X  = (12, 12)  and Y  = (15,6)

And  we can call segment XY the base of triangle CXY...and this is

sqrt [ (15 -12)^2 + ( 12 - 6)^2 ] = sqrt [ 3^2 + 6^2] = sqrt (45) =  3sqrt (5) units in length = (1)

And the equation of the line passing through these two points  is given by

y = -2(x -12) + 12 ...  y  = -2x + 36  or      2x + y - 36  = 0   = AX + By + C  = 0

Now ....the distance from C to this line is the altitude of CXY and it is given by

altitude =  abs (Ax + By + C) / sqrt (A^2 + B^2)   where   (x,y)  are the coordinates of C  = (0,0)

So we have that 

altitude  = abs ( 2(0) + 1(0)  - 36 ) / sqrt ( 2^2 + 1^2)  =  36 / sqrt (5)  = (2)

So.....the area of  triangle CXY  =  base * altitude / 2 =   (1) * (2) / 2 =

[  3 sqrt(5) * 36 ] / [ 2 * sqrt (5) ] =  3 * 36 / 2  =  108 / 2 = 54 sq  units

Now.....since in the original problem the area of XYZ is specified as 21 sq units....and the area of XYZ triangle in the diagram = 18 sq units...we can use a proportion to find the true area of CXY....so we have....

21 / 18   =  true area of CXY /  54

7 / 6  =  true area of CXY / 54

true area of CXY = 7 * 54 / 6  =   7 * 9  = 63 sq units

Points D, E, and F are the midpoints of sides BC, CA, and AB, respectively, of triangle ABC.
Points X, Y, and Z are the midpoints of EF, FD, and DE, respectively.
If the area of triangle XYZ is 21, then what is the area of triangle CXY?

Areas = ?

\(\begin{array}{|rclcrcl|} \hline \triangle ABC &=& A \\ \triangle DEF &=& \frac14 A \\ \triangle XYZ &=& \frac14 \triangle DEF \\ &=& \frac14 \cdot \frac14 A \\ &=& \frac{1}{16} A = 21 \\ \frac{1}{16} A &=& 21 \\ A &=& 16\cdot 21 \\\\ \triangle CXY &=& \frac{AB}{4} \cdot \frac{h}{2} \quad & | & \sin(A) &=& \frac{h}{ \frac{AC}{2} +\frac{AC}{4} } \\ & & & | & \sin(A) &=& \frac{h}{ \frac34 AC } \\ & & & | & h &=& \frac34 AC \sin(A) \\ \triangle CXY &=& \frac{AB}{4} \cdot \frac{ \frac34 AC \sin(A) }{2} \\ &=& \frac{3}{32} \cdot AB\cdot AC\cdot \sin(A) \quad & | & AB\cdot AC\cdot \sin(A) &=& 2A \\ &=& \frac{3}{32} \cdot 2A \\ &=& \frac{6}{32} A \quad & | & A &=& 16\cdot 21 \\ &=& \frac{6}{32} \cdot 16\cdot 21 \\ &=& \frac{6}{2} \cdot 21 \\ &=& 3 \cdot 21 \\ \triangle CXY &=& 63 \\ \hline \end{array}\)