Bdmo
In $\Delta ABC$, Medians AD and CF intersect at P.Let Q be any point on AC.Construct QM and QN parallel to AD and CF respectively.Now the line joining M and N intersects CF and T and AD at U.We have to prove that area of PTU is $\dfrac18$ of the region formed by the lines AD,CF,QM,QN.
We first note that the region formed by AD,CF,QM,QN is a parallelogram.Let QN intersect AD at K and QM intersect CF at L.Then KQLP is a parallelogram.Then we just have to prove that MN passes through midpoints of KP and LP.This cab be done by considering ratios like NK:KQ=2:1 =LM:LQ.Then let I be the midpoint of KQ and O be the midpoint of LQ.Now by constructing a line parallel to AC from N and I.Now we can show that N,the midpoint of KP and M and O are collinear.Then we show similarly that the midpoint of KP,O and M are collinear.Therefore,MN must intersect the KP and PL at their midpoints and we are done.
But I am hoping for a better way to solve this problem.The above approach is extremely tedious and takes a lot of time to write out properly.Any help will be appreciated.
Let $AK = x$ and $AD = m$. Then, $AK : AP = x : \dfrac{2}{3}m$
$NK : FP = x : \dfrac{2}{3}m$ and $KQ : PC = x : \dfrac{2}{3}m$
∴ $NK : FP = KQ : PC$
That is, by properties of median, $KQ : NK = PC : FP = 2 : 1 ……………….[0]$
Notation:- $[\text{figure}]$ = Area of that figure.
∴ $[KUMQ] : [NKU] = 4 : 1$
i.e. $[NKU] + [\text{pink region}] : [NKU] = 4: 1$
∴ $[\text{pink region}] : [NKU] = 3 : 1…………………………..(1)$
Similarly, $[TLM] + [\text{pink region}] : [TLM] = 4 : 1$
∴ $[\text{pink region}] : [TLM] = 3 : 1……………………………(2)$
(1) and (2) imply $[NKU] = [TLM] …………………………..(3)$
It is quite obvious that $\Delta NKU\cong \Delta TLM ……………..(4)$
(3) and (4) imply $\Delta NKU$ is congruent to $\Delta TLM$
Then, by (0), $TL = HK = \dfrac{1}{2}KQ$ meaning that $T$ is the midpoint of $PL$.
The result then follows.