Proving two lengths to be equal

Question

Given this picture how do you proof that GJ=JA.
I am really stuck with this problem.
I know that you get 4 equilateral triangles, but I do not know how to proceed from here.
P.S. $AJ \parallel FG$.

Answer 1

Well, first of all we can write a general function for a first order equation:

$$\text{y}\left(x\right):=\text{a}\cdot x+\text{b}\tag1$$

Now, we have two line:

• $$ \begin{cases} \text{y}_1\left(0\right)=8=\text{a}_1\cdot0+\text{b}_1\\ \\ \text{y}_1\left(x\right)=0=\text{a}_1\cdot6+\text{b}_1 \end{cases}\space\space\space\space\space\space\space\space\space\space\space\therefore\space\space\space\space\space\space\space\space\space\space\space\begin{cases} \text{a}_1=-\frac{4}{3}\\ \\ \text{b}_1=8 \end{cases}\tag2 $$
• $$ \begin{cases} \text{y}_2\left(2\right)=2=\text{a}_2\cdot2+\text{b}_2\\ \\ \text{a}_1=\text{a}_2=-\frac{4}{3} \end{cases}\space\space\space\space\space\space\space\space\space\space\space\therefore\space\space\space\space\space\space\space\space\space\space\space\begin{cases} \text{a}_2=-\frac{4}{3}\\ \\ \text{b}_2=\frac{14}{3} \end{cases}\tag3 $$

We can write $\text{a}_1=\text{a}_2$ because the lines are parallel.

Now, we can write:

$$\left|\text{GJ}\right|=\text{y}_1\left(0\right)-\text{y}_2\left(0\right)=-\frac{4}{3}\cdot0+8-\left(-\frac{4}{3}\cdot0+\frac{14}{3}\right)=\frac{10}{3}\tag4$$

And:

$$\left|\text{AJ}\right|=\sqrt{\left|\text{BJ}\right|^2+\left|\text{AB}\right|^2}=\sqrt{\left(\text{y}_2\left(0\right)-2\right)^2+2^2}=$$ $$\sqrt{\left(-\frac{4}{3}\cdot0+\frac{14}{3}-2\right)^2+2^2}=\frac{10}{3}\tag5$$

DONE!

Answer 2

You could do it geometrically by noting that $\angle FGA = \angle GAJ$ (alternate angles are equal).
Then note that $IG = BG$ and that $AB = AI$.
$\therefore$ $ABGI$ is a kite.
Note that $AG$ is a diagonal and so it bisects $\angle BGI$.
So $\angle BGA = \angle AGI$.
So $\angle JGA = \angle JAG$.
Thus $GJ = JA$.

Or it can be done by calculating the gradient of $GF$ and the point $A (2, 2)$.
Then you can calculate the coordinates of $J$ and calculate $|GJ|$ and $|JA|$ and show that they are equal.

I personally prefer the first method but use whichever is easier and helps more.

Answer 3

Drop a perpendicular $JX $ from $J $ onto the line $FG $. Note triangles $\triangle GJX $ and $\triangle JAB $ are congruent, as both are right, $\angle JGX=\angle AJB $ and $AB=AI=JX $. Thus, $GJ=JA $ follows directly.