Question

A,B and C are the vertices of one triangle.
A has a coordinate of (4,6)
B has a coordinate of (2,-2)
C has a coordinate of (-2,-4)

D is the midpoint of AB
E is the midpoint of AC

Prove that DE is parallel to BC
You must show each stage of your working out.

Answer 1

Answer:

Step-by-step explanation:

Given: The triangle with coordinate A(4,6), B(2,-2) and C(-2,-4). D is the mid point of AB and E is the mid point of AC.

To prove: DE is parallel to BC.

Construction: Join DE.

Proof: If we prove the basic proportionality theorem that is $\frac{AD}{DB}=\frac{AE}{EC}$, then it proves that DE is parallel to BC.

Now, Mid Point D has coordinates=$(\frac{4+2}{2},\frac{6-2}{2})=(3,2)$ and Mid Point E has coordinates=$(\frac{4-2}{2},\frac{6-4}{2})=(1,1)$

Now, AD= $\sqrt{(4-3)^{2}+(6-2)^{2}}=\sqrt{17}$

DB=$\sqrt{(3-2)^{2}+(2+2)^{2}}=\sqrt{17}$

AE=$\sqrt{(4-1)^{2}+(6-1)^{2}}=\sqrt{34}$

EC=$\sqrt{(1+2)^{2}+(1+4)^{2}}=\sqrt{34}$

Now, $\frac{AD}{DB}=\frac{AE}{EC}$

=$\frac{\sqrt{17}}{\sqrt{17}}=\frac{\sqrt{34}}{\sqrt{34}}=\frac{1}{1}$

Hence, $\frac{AD}{DB}=\frac{AE}{EC}$

Thus, By basic proportionality theorem, DE is parallel to BC.

Answer 2

Step-by-step explanation:

If DE and BC are parallel, then they have the same slope.

We have

A(4, 6), B(2, -2), C(-2, -4)

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$

D is the midpoint of AB, and E is a midpoint of AC.

Calculate the coordinateso fo D and E:

$D\left(\dfrac{4+2}{2},\ \dfrac{6+(-2)}{2}\right)\to D\left(\dfrac{6}{2},\ \dfrac{4}{2}\right)\to D(3,\ 2)\\\\E\left(\dfrac{4+(-2)}{2},\ \dfrac{6+(-4)}{2}\right)\to E\left(\dfrac{2}{2},\ \dfrac{2}{2}\right)\to E(1,\ 1)$

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

Calculate the formula of a DE and BC:

$DE:\\\\m_{DE}=\dfrac{2-1}{3-1}=\dfrac{1}{2}\\\\BC:\\\\m_{BC}=\dfrac{-4-(-2)}{-2-2}=\dfrac{-4+2}{-4}=\dfrac{-2}{-4}=\dfrac{1}{2}$

The slope DE and the slope BC are the same.

Therefore DE is parallel to BC.