Problem
GEO-B2-M09-P018 Three Cevians Through One Point
#18
★★★☆☆ Level 3 of 5
In triangle \(A(0,0)\), \(B(6,0)\), \(C(0,6)\), points \(D\in BC\), \(E\in CA\), \(F\in AB\) are chosen so that \(BD:DC=1:2\), \(CE:EA=3:1\), \(AF:FB=2:3\). Prove by coordinates that lines \(AD\), \(BE\), \(CF\) are concurrent.
Find the coordinates of \(D,E,F\), then intersect two lines and check the third.
By segment division, \(D(4,2)\), \(E(0,\frac32)\), \(F(\frac{12}{5},0)\). Line \(AD\): \(y=\frac{x}{2}\). Line \(BE\): \(y=-\frac{x}{4}+\frac32\). Their intersection is \(x=2\), \(y=1\). Line \(CF\) passes through \(C(0,6)\) and \(F(\frac{12}{5},0)\), so it has equation \(y=6-\frac52x\). Point \((2,1)\) lies on it. Hence the three lines are concurrent.
A coordinate version of a problem that can also be solved by Ceva's theorem.