Prove analytically that the line joining the middle points of two sides of a triangle is parallel to the third side and equal half of its length.
Let A, B, C be the vertices of the triangle plotted on the Cartesian Coordinate plane, so let
$\displaystyle A= (x_a,y_a)$
$\displaystyle B= (x_b,y_b)$
$\displaystyle C= (x_c,y_c)$
lets take the midpoints of BC and AC
$\displaystyle BC_m = \left( {\frac{x_b + x_c}{2}, \frac{y_b + y_c}{2}} \right)$
$\displaystyle AC_m = \left( {\frac{x_a + x_c}{2}, \frac{y_a + y_c}{2}} \right)$
using two-point form, the line joining the two midpoints is
$\displaystyle y - \frac{y_a + y_c}{2} = \frac{ \frac{y_b + y_c}{2} - \frac{y_a + y_c}{2} } {\frac{x_b + x_c}{2} - \frac{x_a + x_c}{2}} \left( {x - \frac{x_a + x_c}{2}} \right) = \frac{y_b + y_c - y_a - y_c}{x_b + x_c - x_a - x_c} \left( {x - \frac{x_a + x_c}{2}} \right)$
$\displaystyle \implies y - \frac{y_a + y_c}{2} = \frac{y_b - y_a}{x_b - x_a } \left( {x - \frac{x_a + x_c}{2}} \right)$
but what is the formula of the line joining AB?
$\displaystyle y - y_a = \frac{y_b - y_a}{x_b - x_a} \left( {x - x_a} \right)$
and they have equal values of slopes.. hence, they are parallel.. QED
Here is one way.
Draw any triangle ABC.
D and E are midpoints of AB and BC, respectively.
Let AB = x; BC = y; CA = z; DE = w
To prove analytically that z = 2w. Using Law of Cosines:
In triangle ABC,
z^2 = x^2 +y^2 -2(x)(y)cosB
cosB = (x^2 +y^2 -z^2) /(2xy) -------**
In triangle DBE,
w^2 = (x/2)^2 +(y/2)^2 -2(x/2)(y/2)cosB
cosB = [(1/4)x^2 +(1/4)y^2 -w^2] /[(1/2)(xy)] ------**
cosB = cosB,
(x^2 +y^2 -z^2) /(2xy) = [(1/4)x^2 +(1/4)y^2 -w^2] /[(1/2)(xy)]
Clear the fractions, multiply both sides by 2xy,
x^2 +y^2 -z^2 = (2/(1/2))[(1/4)x^2 +(1/4)y^2 -w^2]
x^2 +y^2 -z^2 = (4)[(1/4)x^2 +(1/4)y^2 -w^2]
x^2 +y^2 -z^2 = x^2 +y^2 -4w^2]
-z^2 = -4w^2
z^2 = 4w^2
Therefore, z = 2w ---------proven.
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To prove that DE is parallel to AC. Using Law of Cosines again,
In triangle ABC,
y^2 = x^2 +z^2 -2(x)(z)cosA
cosA = (x^2 +z^2 -y^2) /(2xz) -------**
In triangle DBE,
(y/2)^2 = (x/2)^2 +w^2 -2(x/2)(w)cosD
cosD = [(1/4)x^2 +(z/2)^2 -(1/4)y^2] /[2(x/2)(z/2)]
cosD = [(1/4)x^2 +(1/4)z^2 -(1/4)y^2] /[2(1/4)xz]
cosD = [x^2 +z^2 -y^2] /[2xz] -----------------------**
So, cosA = cosD
Hence, angle BAC = angle BDE
Since BD is parallel to BA, then, DE is parallel to AC. -------proven.
Hello, the_rookie07!
Prove analytically that the line joining the midpoints of two sides of a triangle
is parallel to the third side and equal half of its length.
Given any triangle, we can orient it so that one vertex is at the origin
. . and another is on the x-axis.Code:| | B | o(p,q) | * * | * * | D o--------o E | * * | * * - A 0 * * * * * * * * o C - (0,0) (r,0) |
We have the vertices: .$\displaystyle A(0,\,0),\;B(p.\,q),\;C(r,\,0)$
The midpoint of $\displaystyle AB$ is: .$\displaystyle D\left(\frac{p}{2},\:\frac{q}{2}\right)$
The midpont of $\displaystyle BC$ is: .$\displaystyle E\left(\frac{p+r}{2},\:\frac{q}{2}\right)$
The slope of $\displaystyle AC$ is: .$\displaystyle m_{_{AC}} \:=\:\frac{0-0}{r-0} \;=\;0$
The slope of $\displaystyle DE$ is: .$\displaystyle m_{_{DE}} \:=\:\frac{\frac{q}{2} - \frac{q}{2}}{\frac{p+r}{2} - \frac{p}{2}} \;=\;0$
. . Therefore: .$\displaystyle {\color{blue}DE \parallel AC}$
The length of $\displaystyle DE \:=\:\sqrt{\left(\frac{p+r}{2} - \frac{p}{2}\right)^2 + \left(\frac{q}{2} - \frac{q}{2}\right)^2} \;=\;\sqrt{\left(\frac{r}{2}\right)^2} \;=\;\frac{r}{2}$
The length of $\displaystyle AC \:=\:r$
. . Therefore: .$\displaystyle {\color{blue}DE \:=\:\frac{1}{2}AC}$
Hi Rookie, there is an easier solution I think using vectors.
Use soroban's triangle picture and imagine you are walking these journeys.
You can see that:-
de = db + be
But also:-
ac = ad + db + be + ec
But point d bisects ab and point e bisects bc therefore:-
ad = bd
and
be = ec
thus, ac = 2db + 2be
or:-
ac = 2( db + be)
but we already stated that de = db + be
so,
ac = 2bd
therefore:-
bd = 1/2 ac.
I doubt there's an easier way to prove it.