Area of triangle 2

圖片參考：http://f20.yahoofs.com/hkblog/ux4rRlOBHhLnw7YkAOw-_1/blog/ap_20090825113945419.jpg?ib_____DCoKVLZmi

In the figure ,
▲ABC = △ABP + △BCQ + △CAR + ▲PRQ................★
Find AP/AD first :
Let H be a point on EC such that DH // BE ,then
CD/DB = CH/HE
y = CH/HE
y+1 = (CH/HE) + 1 = (CH+HE)/HE = CE/HE
i.e. HE = CE/(y+1);
Since AE/EC = x (given), AE = x * EC ; then
AE/AH = AE/(AE+HE)
AE/AH = (x * EC) / [x * EC + CE/(y+1)]
AE/AH = x / [x + 1/(y+1)] =
In △ADH,
AP/AD = AE/AH
We have AP/AD = x / [x + 1/(y+1)]
Find △ABP :
△ABP = △ABD * AP/AD
△ABP = ▲ABC * BD/(BD+DC) * AP/AD ,since CD/BD = y/1,(given)
△ABP = ▲ABC * 1/(1+y) * AP/AD
△ABP = ▲ABC * 1/(1+y) * x/[x + 1/(y+1)]
△ABP = ▲ABC * x / [x(1+y) + 1]
We can find △BCQ and △CAR by above method, but we can write the result directly :
△BCQ = ▲ABC * z / [z(1+x) + 1]
△CAR = ▲ABC * y / [y(1+z) + 1]
Sub the above result to ★:
▲ABC
=▲ABC*{x / [x(1+y)+ 1] + z / [z(1+x)+ 1] + y / [y(1+z)+ 1]}+▲PRQ
Hence :
▲PRQ =
▲ABC * {1 - x / [x(1+y)+ 1] - z / [z(1+x)+ 1] - y / [y(1+z)+ 1]}