# Math Help - Plz Help me with this triangle proof

1. ## Plz Help me with this triangle proof

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.

2. Originally Posted by the_rookie07
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

$A= (x_a,y_a)$
$B= (x_b,y_b)$
$C= (x_c,y_c)$

lets take the midpoints of BC and AC

$BC_m = \left( {\frac{x_b + x_c}{2}, \frac{y_b + y_c}{2}} \right)$

$AC_m = \left( {\frac{x_a + x_c}{2}, \frac{y_a + y_c}{2}} \right)$

using two-point form, the line joining the two midpoints is

$y - \frac{y_a + y_c}{2} = \frac{ \frac{y_b + y_c}{2} - \frac{y_a + y_c}{2} } {\frac{x_b + x_c}{2} - \frac{x_a + x_c}{2}} \left( {x - \frac{x_a + x_c}{2}} \right) = \frac{y_b + y_c - y_a - y_c}{x_b + x_c - x_a - x_c} \left( {x - \frac{x_a + x_c}{2}} \right)$

$\implies y - \frac{y_a + y_c}{2} = \frac{y_b - y_a}{x_b - x_a } \left( {x - \frac{x_a + x_c}{2}} \right)$

but what is the formula of the line joining AB?

$y - y_a = \frac{y_b - y_a}{x_b - x_a} \left( {x - x_a} \right)$

and they have equal values of slopes.. hence, they are parallel.. QED

3. hey thanks alot Kalagota...

4. Originally Posted by the_rookie07
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.
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.

---------------------------------------------
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.

5. Originally Posted by ticbol
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.

---------------------------------------------
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.
this is a nice proof!
but i assume s/he doesn't know what laws of cosine are yet.. Ü

6. Originally Posted by the_rookie07
hey thanks alot Kalagota...
i forgot how to show the distance thing..
i assume you know what to do.. (just use distance formula on A and B and on the midpoints..)

7. hey guys you have been of great help! thanks alot...u guys are like superheroes saving the world at one geometric problem at a time...thanks

8. 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: . $A(0,\,0),\;B(p.\,q),\;C(r,\,0)$

The midpoint of $AB$ is: . $D\left(\frac{p}{2},\:\frac{q}{2}\right)$

The midpont of $BC$ is: . $E\left(\frac{p+r}{2},\:\frac{q}{2}\right)$

The slope of $AC$ is: . $m_{_{AC}} \:=\:\frac{0-0}{r-0} \;=\;0$

The slope of $DE$ is: . $m_{_{DE}} \:=\:\frac{\frac{q}{2} - \frac{q}{2}}{\frac{p+r}{2} - \frac{p}{2}} \;=\;0$

. . Therefore: . ${\color{blue}DE \parallel AC}$

The length of $DE \:=\:\sqrt{\left(\frac{p+r}{2} - \frac{p}{2}\right)^2 + \left(\frac{q}{2} - \frac{q}{2}\right)^2} \;=\;\sqrt{\left(\frac{r}{2}\right)^2} \;=\;\frac{r}{2}$

The length of $AC \:=\:r$

. . Therefore: . ${\color{blue}DE \:=\:\frac{1}{2}AC}$

9. hey thanks Soroban, nice n simple u make that look easy

10. 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:-
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.