Relationship Between Triangles

In triangle $ABC$, segment $\overline{AG}$ from vertex $A$ meets opposite side $\overline{BC}$ at a point $G$ with the length of $\overline{CG}$ 1/3 the length of $\overline{BC}$. That is, point $G$ is located 1/3 of the way along segment $\overline{BC}$ from C. Similarly, points $E$ and $F$ are located at 1/3 marks along the other two sides of triangle $ABC$. The intersections of these segments are the vertices of another triangle $IJH$ in the interior of triangle $ABC$.
What is the relationship between the areas of triangles $ABC$and $IJH$?
Source: NCTM Problem to Ponder 11/16/2010

SOLUTION

Part I

$area AGC=\frac{1}{3}area ABC$  because $GC=\frac{1}{3}BC$
$area BFA=\frac{1}{3}area ABC$   because $AF=\frac{1}{3}AC$
$area CEB=\frac{1}{3}area ABC$   because $EB=\frac{1}{3}AB$
$area AGC+area BFA+area CEB=area ABC\qquad\left (1\right )$
Looking at the figure we notice that, when we are adding areas, the area of triangle $IJH$ is completely left out and that the areas of the three shaded triangles $AHF,BIE$, and $CJG$ are each counted twice. In order for Eq. (1) to be satisfied, we conclude that
$area AHF+area BIE+area CJG=area IJH\qquad(2)$

Part II
Draw $\overline{HD}$ parallel to $\overline{CE}$, $\overline{IK}$ parallel to $\overline{AG}$ and $\overline{JL}$ parallel to $\overline{BF}$.

Triangle $AHF$ is similar to triangle $AJL$ with a scale factor of $\frac{AF}{AL}=\frac{1}{2}$
Triangle $BIE$ is similar to triangle $BHD$ with a scale factor of $\frac{BE}{BD}=\frac{1}{2}$
Triangle $CJG$ is similar to triangle $CIK$ with a scale factor of $\frac{CG}{CK}=\frac{1}{2}$

By property of areas of similar triangles
$\frac{area AHF}{area AJL}=\frac{area BIE}{area BHD}=\frac{area CJG}{area CIK}=\left (\frac{1}{2}\right )^2=\frac{1}{4}$

By property of proportion
$\frac{area AHF+area BIE+area CJG}{area AJL+area BHD+area CIK}=\frac{1}{4}$

Substitute the value of the numerator from Eq. (2)
$\frac{area IJH}{area AJL+area BHD+area CIK}=\frac{1}{4}$

$area AJL+area BHD+area CIK=4\left (area IJH\right )\qquad(3)$

Part III
Draw a parallel line from vertex $I$ to $\overline{AG}$ and a second parallel line from vertex $A$ to $\overline{BF}$. The two parallel lines intersect at point $M$.
Draw a parallel line from vertex $J$ to $\overline{BF}$ and a second parallel line from vertex $B$ to $\overline{CE}$. The two parallel lines intersect at point $N$.
Draw a parallel line from vertex $H$ to $\overline{CE}$ and a second parallel line from vertex $C$ to $\overline{AG}$. The two parallel lines intersect at point $P$.

The constructions create congruent parallelograms and triangles (in the interior of parallelograms) that have equal area because they are halves of congruent parallelograms. For example,
$area AMH=area IHM=area IJH$
$area BNI=area JIN=area IJH$
$area CPJ=area HJL=area IJH$
Let $a=area AMH; b=area BNI; c=area CPJ$
We want to show that $a+b+c=2\left (area IJH\right)$
Triangle $CPL\cong$ triangle $AHF$ by ASA
$\angle {HAF}\cong\angle{LCP}\:\:$ alternate interior angles
$AF=LC=\frac{1}{3}AC$
$\angle{HFA}\cong\angle {PLC}\:\:$ alternate exterior angles
Similarly,
Triangle $BNK\cong$ triangle $CJG$
Triangel $AMD\cong$ triangle $BIE$
We have
$area AMH=area IJH$
$a+area AMD=area IJH$
$a+area BIE=area IJH\qquad\left (4\right )$

$area BNI=area IJH$
$b+area BNK=area IJH$
$b+area CJG=area IJH\qquad\left (5\right )$

$area CPJ=area IJH$
$c+area CPL=area IJH$
$c+area AHF=area IJH\qquad\left (6\right )$

Adding equations (4), (5), and (6)
$a+b+c+\left (area BIE+area CJG+area AHF\right )=3\left (area IJH\right )$
$a+b+c+\left (area IJH\right )=3\left (area IJH\right )$    by Eq. (2)
$a+b+c=2\left (area IJH\right )\qquad\quad\left (7\right )$

Summary

We have decomposed the entire triangle $ABC$ into 7 distinct and contiguous triangles the areas of which are related as follows:
$area IJH=area IJH$
$area AJL+area BHD+area CIK=4\left (area IJH\right )$ by Eq. (3)
$area ADH+area BKI+area CLJ=2\left (area IJH\right )$ by Eq. (7)
The ratio $area IJH\: :\:area ABC$ is $1\: :\:7$.

Answer: $1\: :\:7$.