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
lets take the midpoints of BC and AC
using two-point form, the line joining the two midpoints is
but what is the formula of the line joining AB?
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: .
The midpoint of is: .
The midpont of is: .
The slope of is: .
The slope of is: .
. . Therefore: .
The length of
The length of
. . Therefore: .
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.