Question 1197198: Let ABCD be a rectangle and M a point inside it.
Prove that (MA)^2 + (MC)^2 = (MB)^2 + (MD)^2
Found 2 solutions by math_tutor2020, ikleyn:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
This is what the diagram looks like
We have rectangle ABCD. Inside the rectangle is a point M that can freely float around. Place it anywhere you want.
Segments MA, MB, MC, and MD connect M with each corner point A,B,C,D.
The claim is that
(MA)^2 + (MC)^2 = (MB)^2 + (MD)^2
After thinking through it a bit, notice how both sides involve sum of squares.
Think through various algebra lessons/lectures in the past.
Where have you seen a sum of squares? Hopefully the pythagorean theorem comes to mind as this idea comes up a lot in math topics.
That theorem says (legA)^2 + (legB)^2 = (hypotenuse)^2
We'll need to involve the pythagorean theorem somehow, which means we need right triangles.
Draw a vertical line through point M so we form points E and F as shown below.
Point E is on segment AB.
Point F is on segment CD.
Focus on triangle AEM. It is a right triangle which allows us to be able to use the pythagorean theorem
(AE)^2 + (EM)^2 = (MA)^2
Similar logic will apply for the other triangles I'll mention.
Now move onto triangle MFC
Use the pythagorean theorem to get
(MF)^2 + (FC)^2 = (MC)^2
So,
(MA)^2 + (MC)^2 = [ (AE)^2 + (EM)^2 ] + [ (MF)^2 + (FC)^2 ]
(MA)^2 + (MC)^2 = (AE)^2 + (EM)^2 + (MF)^2 + (FC)^2
I'll color code the horizontal sides as red and the vertical sides as blue to get this
(MA)^2 + (MC)^2 = (AE)^2 + (EM)^2 + (MF)^2 + (FC)^2
Keep these terms in mind for later when we compare them to the terms of (MB)^2 + (MD)^2.
Now let's move to triangle BEM
Once again we use the pythagorean theorem to get...
(EB)^2 + (EM)^2 = (MB)^2
But wait, notice that EB = FC since rectangle EBCF has congruent opposite sides.
This allows us to replace EB with FC
(EB)^2 + (EM)^2 = (MB)^2
(FC)^2 + (EM)^2 = (MB)^2
Next, move onto triangle DFM
Again we use the pythagorean theorem.
(DF)^2 + (MF)^2 = (MD)^2
and then replace DF with AE due to rectangle AEFD having congruent opposite sides
So we have
(DF)^2 + (MF)^2 = (MD)^2
(AE)^2 + (MF)^2 = (MD)^2
-----------------------------------------------
In short,
(FC)^2 + (EM)^2 = (MB)^2
(AE)^2 + (MF)^2 = (MD)^2
which sum to
(MB)^2 + (MD)^2 = [ (FC)^2 + (EM)^2 ] + [ (AE)^2 + (MF)^2 ]
(MB)^2 + (MD)^2 = (FC)^2 + (EM)^2 + (AE)^2 + (MF)^2
(MB)^2 + (MD)^2 = (AE)^2 + (EM)^2 + (FC)^2 + (MF)^2
(MB)^2 + (MD)^2 = (AE)^2 + (EM)^2 + (MF)^2 + (FC)^2
In step 3 I swapped the positions of FC and AE
In step 4 I swapped the positions of FC and MF
Hopefully you notice that the result of (MB)^2 + (MD)^2 is exactly identical to the result of (MA)^2 + (MC)^2. Carefully check each term to verify.
Therefore we have proven that if M is anywhere inside rectangle ABCD, then (MA)^2 + (MC)^2 = (MB)^2 + (MD)^2
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
Imagine that the rectangle ABCD lies in the first quadrant;
that its vertex A is the origin of the coordinate system,
vertex B lies on x-axis and vertex D lies on y-axis.
Let x-dimension of the rectangle be "a", and
y-dimension of the rectangle be "b".
Let point M has coordinates M = (x,y).
Then (MA)^2 + (MC)^2 = (x^2 + y^2) + ((a-x)^2 + (b-y)^2). (1)
(MB)^2 + (MD)^2 = ((a-x)^2 + y^2) + (x^2 + (b-y)^2). (2)
You can see at a first glance that expressions (1) and (2) are equal, since they are identical
(the difference is only in the order of addends).
So, the problem is just solved and the required equality is proved.
Solved and completed.
|
|
|
