How can i Find the Area of a Traingle Formed inside a Triangle?

Question

Here we have Co Ordinates of A ,B ,C and We have to Find the Area of the Triangle PQR Formed Inside the Triangle ABC.triangle ABC. Point D, E and F divides the sides BC, CA and AB into ratio 1:2 respectively. That is CD=2BD, AE=2CE and BF=2AF. A, D; B, E and C, F are connected. AD and BE intersects at P, BE and CF intersects at Q and CF and AD intersects at R. Please Check the Image in the link.

Answer 1

Denote by $[ABC]$ the area of the triangle $\Delta ABC$.

First note that $$\frac{[ADC]}{[ABC]}=\frac{\overline{DC}}{\overline {BC}}=\frac{2}{3}\implies [ADC]=\frac{2[ABC]}{3}\tag{1}\label{1}$$

Now, by Menelaus's Theorem regarding the triangle $\Delta ADB$ and the line through $F,R$ and $C$: $$\frac{\overline {AF}}{\overline {FB}}\cdot\frac{\overline {BC}}{\overline {DC}}\cdot\frac{\overline {RD}}{\overline {AR}}=1 \implies \frac{\overline {RD}}{\overline {AR}}=\frac{\overline {FB}}{\overline {AF}}\cdot\frac{\overline {DC}}{\overline {BC}}=2\cdot\frac{2}{3}=\frac{4}{3}$$

Note now that (recall $\eqref{1}$) $$\frac{\overline {RD}}{\overline {AR}}=\frac{[ADC]-[ARC]}{[ARC]}=\frac{[ADC]}{[ARC]}-1=\frac{2[ABC]}{3[ARC]}-1=\frac{4}{3}$$ $$\therefore [ARC]=\frac{2[ABC]}{7}$$

Analogously $$[ARC]=[CQB]=[BPA]=\frac{2[ABC]}{7}$$ Since $$[PQR]=[ABC]-\Bigl([ARC]+[CQB]+[BPA]\Bigr)$$ Your have $$[PQR]=[ABC]-3\Biggl(\frac{2[ABC]}{7}\Biggr)=\frac{[ABC]}{7}$$

$\mathbf {Remark:}$

The problem (configuration) you've suggested is known as the "One-seventh area triangle" and is a special case of "Routh's theorem" which gives a general solution in order to calculate the area of the inner triangle for any three cevians forming this one.