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 ABCand 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.

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , , . Bookmark the permalink.

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