Area of a triangle The following problem in elementary geometry was proposed to me. As a mathematical analyst, I confess that I can't solve it. And I have no idea of what I could do. Here it is: pick a triangle, and draw the three mediana (i.e. the segments that join a vertex with the midpoint of the opposite side). Use the three segments to construct a second triangle, and prove that the area of this triangle is $3/4$ times the area of the original triangle.
Any help is welcome.
 A: I believe there is a nice visual explanation for this.
Start with a triangle and draw its medians.

Then construct congruent triangles to form a hexagon, and connect every other vertex on its perimeter.

Let $A=$ (area of the original triangle) and $B=$ (area of the triangle of medians). The shaded triangle has area $4B$, because its sides are twice as long as the medians, and it is also equal to half the area of the hexagon, which is $6A/2=3A$. $3A=4B$, so $B=\frac{3}{4}A$.
A: As suggested we can define any triangle with two vectors: $\mathbf{a}=\begin{pmatrix}{a_1 \\a_2} \end{pmatrix}$ and $\mathbf{b} = \mathbf{e}_1$, such that $\mathbf{a}$ and $\mathbf{b}$ are not colinear and where $\mathbf{b}$ has been chosen for simplicity. 
Then a linear map, $A$, can be constructed to send these vectors to there corresponding line segments. If we get that the determinant of $A$ is such that $\det{A}=\frac{3}{4}$ the result will be proved.
After drawing a picture of a triangle defined in such a way it is clear that we want the map $A$ such that: 
$$\mathbf{a} \mapsto \mathbf{b} - \frac{1}{2}\mathbf{a}  , $$
$$\mathbf{b} \mapsto \mathbf{a} - \frac{1}{2}\mathbf{b}   .$$
This completely determines $A$ and after solving some equations we get that: $$A=\begin{pmatrix}1-\frac{a_1}{2} & -\frac{1+a_1^2}{2a_2}\\-\frac{a_2}{2} & 1+\frac{a_1}{2} \end{pmatrix}.$$
Upon computing $\det(A)$ we get that it is $\frac{3}{4}$ proving what is required.
A: Use vectors.It will be really helpful.Define sides of triangle ABC as A=0, B= b vector and C= c vector.
A: In a triangle $ABC$ with medians intersecting at $O$, draw a line throgh $A$ parallel to the median through $B$ and a line through $B$ parallel to the median through $A$. Let $D$ be the intersection of the new lines. Then the parallelogram $AOBD$ has area $2/3$ of the area of the triangle $ABC$ but it can also be partitioned by the diagonal $OD$ into two triangles made out of intervals of lengths $2/3$ of the lengths of the medians themselves. Hence the area of the triangle formed by the medians is $(3/2)^2$ times $1/3$, or $3/4$ of the area of the triangle $ABC$.
A: 
If you don't understand vector that well then you can use this proof
Assume that three medians BD,AE and CF are p,q,r respectively and the triangle made by the medians is $\triangle {X}$
Now we draw parallelogram $ GCIA $. Area of $ \triangle {GAC}= \frac {1}{3} \triangle {ABC} $ ( As the medians divide a triangle into six congruent triangles. And there is two of them in $ \triangle {GAC} $ ).
Now $\triangle {GAI} = \triangle {GAC} $
$ \triangle {GAC}= \frac {1}{3} \triangle {ABC} $
$ GA= \frac {2}{3} AE = \frac {2}{3} q $
similarly $GI = \frac {2}{3} p$ and $AI= \frac {2}{3}r$
So $\triangle {GAI} $ is triangle whose sides are two third of $\triangle {X}$.
So the area of $\triangle {GAI} = \frac {4}{9} \triangle {X}$
$ \frac {4}{9} \triangle {X} = \frac {1}{3} \triangle {ABC}$
$ \frac {4}{3} \triangle {X} = \triangle {ABC} $
A: $\newcommand{\+}{^{\dagger}}%
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Let's $\vec{a}$, $\vec{b}$ and $\vec{c}$ the vertexs of the original triangle. Its area ${\cal A}$ is given by the magnitude of:
$$
\half\,\vec{a}\times\vec{b} + \half\,\vec{b}\times\vec{c} + \half\,\vec{c}\times\vec{a}
$$ 
Middle points of the triangle are: 
$$
\half\,\pars{\vec{a} + \vec{b}}\,,\ \half\,\pars{\vec{b} + \vec{c}}\ \mbox{and}\ \half\,\pars{\vec{c} + \vec{a}}
$$
The 'new' area is given by:
\begin{align}
&\half\bracks{\half\,\pars{\vec{a} + \vec{b}}\times\half\,\pars{\vec{b} + \vec{c}}}
+
\half\bracks{\half\,\pars{\vec{b} + \vec{c}}\times\half\,\pars{\vec{c} + \vec{a}}}
+
\half\bracks{\half\,\pars{\vec{c} + \vec{a}}\times\half\,\pars{\vec{a} + \vec{b}}}
\\[3mm]&=
{\vec{a}\times\vec{b} + \vec{a}\times\vec{c} + \vec{b}\times\vec{c} \over 8}
+
{\vec{b}\times\vec{c} + \vec{b}\times\vec{a} + \vec{c}\times\vec{a} \over 8}
+
{\vec{c}\times\vec{a} + \vec{c}\times\vec{b} + \vec{a}\times\vec{b} \over 8}
\\[3mm]&={\vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a} \over 8}
\\[3mm]&=\color{#00f}{\Large{1 \over 4}}\,\pars{{\vec{a}\times\vec{b} + \vec{b}\times\vec{c} + \vec{c}\times\vec{a} \over 2}}
\end{align}
A: Let $AA', BB', CC'$ be the medians, and let $G$ be their intersection. Let $G'$ be the symmetric of $G$ with respect to $A'$. Then, since $BGCG'$ is a parallelogram, in the triangle $BGG'$ we have
$$BG=\frac{2}{3}BB' \,;\,  BG'=\frac{2}{3}CC' \,;\, GG'=\frac{2}{3}AA' \,.$$
Thus, area of $BGG'$ is $\frac{4}{9}$ area of the triangle made by the medians.
Also, area $BGG'=\frac{1}{2}$ area $BGCG'$=area $BGC=\frac{1}{3}$ area $ABC$.
[the last equality follows either by repeating the argument by the other two sides, or observing that $ABC$ and $GBC$ have the same base, and their heights are $3:1$].
Thus
$$\mbox{Area} BGG'=\frac{4}{9} \mbox{Area of the median triangle} $$
$$\mbox{Area} BGG'=\frac{1}{3} \mbox{Area} ABC$$
Simplification Here is a much simpler version of this proof:
Note the the area of $BGC$ is $\frac{1}{3}$ area of $ABC$. 
Now cut  $BGC$ along the side $GA'$ and rotate $CGA'$ by $180^o$ degrees. Then you get a triangle with the sides $\frac{2}{3}$ the medians. Equate the areas.
A: "Here it is: pick a triangle":   
Since we get to pick the triangle, I will pick a triangle T1 that has equal sides, therefore 3 equal angles = 60 degrees.  Let's call the sides "A".   The area of this triangle T1 is equal to [0.5 * A * A].  Each of the 3 "medianas" will have a magnitude equal to [A * Sin(60)], therefore a triangle T2 constructed from these 3 medianas will have an area equal to (0.5 * [A * Sin(60)] * [A * Sin(60)])
Dividing the area of Triangle T2 by the area of triangle T1:
Ratio of Areas = (0.5 * [A * Sin(60)] * [A * Sin(60)]) / [0.5 * A * A]
Ratio of Areas = Sin(60)  *  Sin(60)
Ratio of Areas = 0.75 = 3/4
