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If triangle $ABC$ is equilateral,$BD/BC=1/3, CE/CA=1/3,$ and $ AF/AB=1/3$. What is the ratio of the area of triangle? I have problems analyzing this triangle I tried to use phythagorean, heron's formula, the formula $A$=$\frac{ab\sin(\theta)}{2}$ but I still can't figure the inside area. I've also done some research they said it's $1:7$? But what I want is to understand how they got $ 1:7$.

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  • $\begingroup$ There's no D, E, and F in your diagram. (And this appears to be plane rather than solid geometry). $\endgroup$ Nov 28, 2014 at 1:48
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    $\begingroup$ See Routh's theorem. $\endgroup$
    – Lucian
    Nov 28, 2014 at 2:41

2 Answers 2

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There are many proofs of this result (see here for a generalized version), but perhaps the easiest is using the theorem of Menelaus, along with some area ratios.

enter image description here

We use this figure. Consider collinear points $A, K, E$ on the sides of $\triangle DBG$. Menelaus's Theorem tells us:

$$\begin{align} \frac{DA}{AB} \frac{BK}{KG} \frac{GE}{ED} &= 1 \\ \frac{3}{1} \frac{BK}{KG} \frac{2}{1} &= 1\\ \implies \frac{BK}{KG} &= \frac{1}{6}\\ \implies \frac{KG}{BG} &= \frac{6}{7} \end{align}$$

Thus

$$\begin{align} [\triangle KAG] &= \frac{KG}{BG} [\triangle BAG] \\ &= \frac{6}{7} \frac{BA}{DA}[\triangle DAG]\\ &= \frac{6}{7} \frac{1}{3} [\triangle DAG]\\ &= \frac{6}{21} [\triangle DAG] \end{align}$$

By symmetry, we have that

$$ [\triangle KAG] = [\triangle LDA] = [\triangle MGD] = \frac{6}{21}[\triangle DAG] $$

Thus,

$$ \begin{align} [\triangle LKM] &= [\triangle DAG] - ([\triangle KAG] + [\triangle LDA] + [\triangle MGD])\\ &= ( 1 - \frac{6}{21} - \frac{6}{21} - \frac{6}{21})[\triangle DAG])\\ &= \frac{1}{7}[\triangle DAG] \end{align}$$

As desired.

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  • $\begingroup$ how did $\frac{GE}{ED}$ become $\frac{2}{1}$? $\endgroup$
    – Mickey
    Dec 3, 2014 at 1:11
  • $\begingroup$ @Mickey Remember that the cevians are trisecting the opposite sides. $\endgroup$ Dec 3, 2014 at 2:44
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Borrowing @extremeaxe5's diagram. If you don't know Menelaus' theorem, you can deduce $\frac{KG}{BG}=\frac67$ using vector geometry:

By hypothesis $B=\frac23A+\frac13D$ and $E=\frac23D+\frac13G$. Point $K$ lies on the line through $AE$, so it satisfies the equation $$\textstyle K=A+\alpha(E-A)=(1-\alpha)A +\frac23\alpha D+\frac13\alpha G\tag1$$ for some $\alpha$, and $K$ also lies on the line through $GB$, so $K$ also satisfies $$\textstyle K=G+\beta(B-G)=\frac23\beta A+\frac13\beta D+(1-\beta)G\tag2 $$ for some $\beta$. Solve (1) and (2) simultaneously for $\alpha=\frac37$ and $\beta=\frac67$. In particular (2) says $K = G+\frac 67(B-G)$, which means $K$ is $\frac67$ of the way from $G$ to $B$.


From here you can continue as in @extremeaxe5's proof. An alternative is to observe the inner triangle is equilateral by symmetry, so it is enough to determine the length of any one of its sides. Plugging $\beta=\frac67$ into (2) yields $ K=\frac47 A + \frac27 D+\frac17 G $. A similar argument gets $L=\frac47 D+\frac27 G+\frac17 A$. Subtract these: $$\textstyle K-L=\frac37A -\frac27 D-\frac17G=\frac37(A-D)-\frac17(G-D) $$ and compute the squared length: $$ \textstyle |K-L|^2=(\frac37)^2|A-D|^2+(\frac17)^2|G-D|^2 -2\cdot\frac37|A-D|\cdot\frac17|G-D|\cdot\cos\theta $$ where $\theta=60^\circ$ is the angle between vectors $A-D$ and $G-D$. Simplify, using the fact that the outer triangle is equilateral, to obtain $$\textstyle|K-L|^2=\frac17|A-D|^2. $$

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