Geometry problem.

[Josie]

I'm absolutely terrible at proofs or anything like that, just bad at Geometry in general.
My teacher is a nice lady and may help a few other students, but she goes through everything so quickly and I'm just lost.
Anyways the problem says,

Prove that ALEX is a square. A(-5,5) L(7,10) E(12,-2) X(0,-7)

So I graphed it with no problem, and took down the slope, because I'm guessing that's what i'm supposed to do?
I got 5/12 for all of them, which would make sense because they're all supposed to be the same because it's a square and they're even.
So that's all I have,
I graphed it and got the slope, but when it comes to writing the proofs; I'm lost. All I have is the t-chart that says -statement- and -reason-
Please help?

 

[srmichael]

I'm absolutely terrible at proofs or anything like that, just bad at Geometry in general.
My teacher is a nice lady and may help a few other students, but she goes through everything so quickly and I'm just lost.
Anyways the problem says,

Prove that ALEX is a square. A(-5,5) L(7,10) E(12,-2) X(0,-7)

So I graphed it with no problem, and took down the slope, because I'm guessing that's what i'm supposed to do?
I got 5/12 for all of them, which would make sense because they're all supposed to be the same because it's a square and they're even.
So that's all I have,
I graphed it and got the slope, but when it comes to writing the proofs; I'm lost. All I have is the t-chart that says -statement- and -reason-
Please help?

That statement is incorrect. What makes a quadrilateral a square? All of the lengths are the same AND all four angles are 90º. So to prove this is a square you have to:

1) Show that all the lengths are the same
2) All the angles are 90º

For 1), you can use the distance formula. Do you know that formula?
For 2), two segments intersect at 90º if they are perpendicular. Lines that are perpendicular have slopes that are negative reciprocals of each other. For example, if one slope is -3/5 the slope of the segment perpendicular to it would be 5/3. Get it? Another way of thinking about this is that the products of the slopes of perpendicular lines must be -1 (-3/5)(5/3) = -1

So try and do 1 and 2 and come back to us with your findings.

 

[soroban]

Hello, Josie!

\(\displaystyle \text{Prove that }ALEX\text{ is a square: }\:A(\text{-}5,5),\;L(7,10),\;E(12,\text{-}2),\;X(0,\text{-}7)\)

              |     L
              |     o(7,10)
              | *    *
        A   * |       *
        o     |        *
   (-5,5)*    |         *
    - - - * - + - - - - -* - - - - - -
           *  |           o(12,-2)
            * |       *   E
             *|   *
           X  o(0,-7)
              |

We must show that the four sides have equal length.
. . Hence, we have a rhombus.
Then prove that one vertex has a right angle.


\(\displaystyle \begin{array}{cccccccccc}\overline{AL} &=& \sqrt{(7-[\text{-}5])^2 + (10-5)^2} &=& \sqrt{12^2 + 5^2} &=& \sqrt{169} &=& 13 \\ \overline{LE} &=& \sqrt{(12-7)^2 + (\text{-}2-10)^2} &=& \sqrt{5^2 + 12^2} &=& \sqrt{169} &=& 13 \\ \overline{EX} &=& \sqrt{(0-12)^2+(-7-[\text{-}2])^2} & =& \sqrt{12^2+5^2} &=& \sqrt{169} &=& 13 \\ \overline{XA} &=& \sqrt{(\text{-}5-0)^2 + (5-[\text{-}7])^2} &=& \sqrt{5^2+13^2} &=& \sqrt{169} &=& 13 \end{array}\)



\(\displaystyle m_{_{AL}} \:=\:\dfrac{10-5}{7-(\text{-}5)} \:=\:\dfrac{5}{12}\)

\(\displaystyle m_{_{LE}} \:=\:\dfrac{\text{-}2-10}{12-7} \:=\:\text{-}\dfrac{12}{5}\)


\(\displaystyle \text{Since }\,m_{_{AL}}\!\!\cdot\!m_{_{LE}} \:=\: \left(\tfrac{5}{12}\right)\left(\text{-}\tfrac{12}{5}\right) \:=\:\text{-}1,\;\text{ then: }\:\overline{AL} \perp \overline{LE}.\)


\(\displaystyle \text{Q.E.D.}\)

 

[lookagain]

I'm looking for an alternative to prove it's a square.


For simpler notation, let AL, LE, EX, and XA represent the lengths
of those respective line segments, as well as the name of the line
segments themselves.

Look at two pairs of opposite corner vertices, A & E, and also L & X.

Their respective line segments are AE and LX.

They are diagonals of the quadrilateral, and then hopefully at the end
of the work will, in addition, be determined to be diagonals of a square.


1) Establish that the midpoints of AE and LX are the same point.


2) Establish that diagonal AE is congruent to diagonal LX by using the
squares of distances from their respective endpoints. ( Don't bother
working with the actual distances.That is, don't take square roots.)


3) Establish that any pair of consecutive sides are congruent to each other.




These three parts worked out together should be sufficient to
show that ALEX is a square.

 

[soroban]

Hello, Denis

Agree; but would it not be easier (once you know sides = 13)
to calculate a diagonal's length; then is it equal to SQRT(13^2 + 13^2)?

I had thought of various other methods and none of them seemed simple enough.

If we show that a diagonal is \(\displaystyle 13\sqrt{2}\),
then we must explain why that forces the figure to be a square.

I thought of showing that the two diagonals are equal.
But that too requires an explanation.

 

[lookagain]

The following procedure should be even more streamlined than the other methods (read: fewer steps).

Background (in the context of the square shapes):


The length of a diagonal equals \(\displaystyle \sqrt{2}\) times the length of a side.

Then the square of the length of a diagonal equals twice the square of the length of a side.


Let s = the length of the side of a square.

Let d = the length of the diagonal of that square


\(\displaystyle d \ = \ (\sqrt{2})s \ \implies\)

\(\displaystyle d^2 \ = \ 2s^2\)


The midpoints of a square's diagonals coincide.


Outline of the proof:

1) Show that the square of the length of a diagonal equals twice the square of the length of any of the sides (line segments).

2) Show that the squares of the lengths of the diagonals equal each other.

3) Show that the midpoints of the diagonals coincide (have the same set of co-ordinates).

 

[Deleted member 4993]

I seem to remember that there is a theorem which states:

A quadrilateral with four equal sides and one internal angle equal to right angle is a square.

I remember constructing squares (with compass and rulers) using this theorem (or corollary to some theorem).

So - show four sides are equal and (one diagonal length)² = sum (length of two adjacent sides)² and you are done.