Special Medians of a Triangle

Draw an extra line between $DE$

Triangle $ADE$ is similar to Triangle $ABC$

If we say $BC$ is $x$ , means $DE$ is $\frac x2$

We can say :

$x^2$ = $BG^2$ + $CG^2$

$(\frac x2)^2$ = $DG^2$ + $EG^2$

Adding them up would be :

$\frac 54x^2$ = $(BG^2 + EG^2)$ + $(CG^2 + DG^2)$

$\frac 54x^2$ = $(\frac 72)^2$ + $(\frac 92)^2$

$\frac 54x^2$ = $\frac {49}{4}$ + $\frac {81}{4}$

$5x^2$ = $49+81$

$x^2$ = $\frac {130}{5}$ = $26$

We know that \overline{BD} and \overline{EC} are medians because they connect the midpoint of a side of the triangle and the opposite vertex. Therefore the centroid (point G) separates \overline{EC} into ratios of 1:2 and \overline{BD} into ratios of 1:2. Set \overline{BG} as 2x, \overline{GD} as x, \overline{CG} as 2y and \overline{GE} as y. By Pythagorean Theorem, x^2 + 4y^2 = (\frac{9}{2})^2 and 4x^2 + y^2 = (\frac{7}{2})^2. We desire 4x^2 + 4y^2 because this will equal (\overline{BC})^2. Adding both equations together leads to 5x^2+5y^2 = \frac{65}{2}. 4x^2+4y^2 = 26.

Let BD be 3x and CE be 3y
So by the property of Centroid,
$\frac {BG}{GD}$ = $\frac {CG}{GD}$ = $\frac {2}{1}$ Also,
$BE^{2}$ = $EY^{2} + BY^{2}$
$\frac {7}{2}^{2}$ = $y^{2} + (2x)^{2}$
$\frac {49}{4}$ = $y^2 + 4x^2$ Eq. No.: 01
Similarly,
$CD^{2}$ = $GD^{2} + CG^{2}$
$\frac {9}{2}^{2}$ = $x^{2} + (2y)^{2}$
$\frac {81}{4}$ = $x^2 + 4y^2$ Eq. No.: 02
Upon adding Eq. 1 and Eq. 2 we get,
$5 \times (x^2 + y^2)$ = $\frac {130}{4}$
Upon multiplying this by $\frac {4}{5}$ we get,
$4x^2 + 4y^2$ = 26
$(2x)^2 + (2y)^2$ = 26
$BG^{2} + GC^{2}$ = 26
$BC^{2}$ = 26

Because angles BGE and CGD are right angles, it can be concluded via the Pythagorean Theorem that $EG^2$ + $BG^2$ = $3.5^2$ , and $CG^2$ + $DG^2$ = $4.5^2$ . Furthermore, $DG^2$ + $EG^2$ = $DE^2$ . Because DE is the midline of triangle ABC (D and E are respective midpoints of the sides they are on), its length must be half that of BC. Finally, $BG^2$ + $CG^2$ = $BC^2$ . Substituting and simplifying: $BC^2$ = $BG^2$ + $CG^2$ , $BC^2$ = $3.5^2$ - $EG^2$ + $4.5^2$ - $DG^2$ , $BC^2$ = 32.5 - ( $EG^2$ + $DG^2$ ), $BC^2$ = 32.5 - $ED^2$ , $BC^2$ = 32.5 - $(BC/2)^2$ , $(5/4)BC^2$ = 32.5, $BC^2$ = 26

Let CG=a, GE=b, GB=c, GD =d, and BC=x. Using Pythagoras, we have a^2 + c^2 = x^2 ...(1), a^2 + d^2 = 4.5^2 .....(2), and b^2 + c^2 = 3.5^2 ........(3). We add (2) and (3) these 3 to get a^2+b^2+c^2+d^2 = 4.5^2+3.5^2. Substitute (1) to this equation. then we have x^2 + b^2+d^2 = 4.5^ + 3.5^. But b^2+d^2 is (DE)^2 which is (0.5 x)^2. So we have 0.25x^2 + x^2 = 32.5, then we find x^2 = (BC)^2 = 26

Noting that the triangles EBG, CGD, GCB and DEG are right triangles we deduce the validity of the following reports:

EB^2-BG^2=GE^2

DC^2-CG^2=GD^2

CG^2+GB^2=CB^2

DG^2+GE^2=DE^2

So:

EB^2-BG^2+DC^2-CG^2=GE^2+GD^2=DE^2

So:

EB^2+DC^2-(BG^2+CG^2)=DE^2,

EB^2+DC^2-CB^2=DE^2.

Observing that:

CB^2=4DE^2 (CB=2DE)

we may write:

EB^2+DC^2-CB^2=(CB^2)/4.

SO:

EB=7/2, CD=9/2

CB^2=26

Medians are divided in 2:1 by centroid, thus

GE=b and GC=2b for some a and b GD=a and GB=2a

By Pythagoras Theorem,

(2a^2) + (b^2) = ((7/2)^2), (2b^2) + (a^2) = ((9/2)^2)

Solving the two for positive a and b, we get

a=(sqrt(23/3))/2 and b=(sqrt(55/3))/2

Again by Pythagoras,

(BC)^2 = (2a)^2 + (2b)^2 = 78/3 = 26

Since $BD$ and $CE$ are medians, thus $AE=EB$ , $AD=DC$ and $DE$ is parallel to $BC$ . It also follows that triangles $ADE$ and $ACB$ are similar, thus $\frac{BC}{DE} = \frac{AB}{AE} = 2$ $\Rightarrow BC = 2DE$ . Since $DE$ is parallel to $BC$ thus $\angle EDB = \angle DBC$ and $\angle DEC = \angle ECB$ , which implies that triangles $DEG$ and $BCG$ are similar by angle-angle-angle and a factor of $\frac{BC}{DE}=2$ .

Let $GD = x$ and $GE = y$ , so we have $BG = 2x$ and $CG = 2y$ . Applying the Pythagorean theorem on triangles $GEB$ , $GDC$ and $GBC$ , we have $\frac{AB^2}{4} = EB^2 = 4x^2 + y^2$ , $\frac{AC^2}{4} = DC^2 = x^2 + 4y^2$ and $BC^2 = 4x^2 + 4y^2$ . Adding the first two equations, we have

$\begin{aligned} \frac{AB^2 + AC^2}{4} &= 5x^2 + 5y^2 \\ &= \frac{5}{4}BC^2 \\ BC^2 &= \frac{AB^2 + AC^2}{5} \\ &= \frac{7^2 + 9^2}{5} \\ &= 26 \\ \end{aligned}$

In triangle $ABC$ , $D$ is the midpoint of $AC$ and $E$ is the midpoint of $AB$ . $BD$ and $CE$ are perpendicular to each other and intersect at the point $G$ .

Because $G$ is median and $BD$ and $CE$ are perpendicular to each other, implies $CG = 2EG$ and $BG = 2DG$ and we got 3 right triangle of $CGD$ , $BGE$ and $BGC$ .

Just do with Pythagorean theorem on each triangle.

Let see in triangle $CGD$

$CG^2 + DG^2 = CD^2$

because $CD=\frac {CA}{2}=\frac {9}{2}$ and $DG=\frac {BG}{2}$

$CG^2 + (\frac {BG}{2})^2 = (\frac {9}{2})^2$

$4CG^2 + BG^2 = 9^2$ .. (1)

in triangle $BGE$

$BG^2 + EG^2 = BE^2$

because $BE=\frac {BA}{2}=\frac {7}{2}$ and $EG=\frac {CG}{2}$

$BG^2 + (\frac {CG}{2})^2 = (\frac {7}{2})^2$

$4BG^2 + CG^2 = 7^2$ ... (2)

and in triangle $BGC$

$BC^2 = CG^2 + BG^2$ ... (3)

Let's sum (1) and (2) we'll got

$5BG^2 + 5CG^2 = 9^2 + 7^2$ divide it with 5

$BG^2 + CG^2 = \frac{9^2 + 7^2}{5}$

and subtitute to (3)

$BC^2 = \frac{9^2 + 7^2}{5} = 26$

BD is a median of the triangle, so BG = 2 DG. Similarly, CG = 2 EG. Let DG = a and EG = b. Then, by Pythagorean theorem on triangle BGE, 4a^2+b^2=(7/2)^2=49/4. By Pythagorean theorem on triangle CGD, a^2+4b^2=(9/2)^2=81/4. Adding gives 5a^2+5b^2=130/4=65/2, so a^2+b^2=13/2.

What we want to find is BC^2 which is 4a^2+4b^2 by Pythagorean theorem again. This is 4*13/2=26.