# Geometric proof

1. Oct 11, 2004

### galois427

the prob ask me to prove that if parallelograms ABCD and AXYZ (see the attachment) are such that point X lies on side BC and point D lies on side YZ, the area of the 2 parallelograms are equal.

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2. Oct 11, 2004

### TenaliRaman

use vectors ....
AX = AB+BX
AZ = AD+DZ
take cross product of AX and AZ and simplify ...

-- AI

3. Oct 11, 2004

### galois427

how would you express AX and AZ as a vector? I just set z=0 but when I took the cross product, i have a bunch of unknowns (x1,x2,x3,y0,y1).

4. Oct 12, 2004

### TenaliRaman

we are working in the 2d plane so all vectors will be of form xi+yj tho we don't need to use this form here.
AX x AZ = (AB+BX) x (AD+DZ)
simplify this!!
your goal is to show AX x AZ = AB x AD
tis a bit tricky but u can do it for sure!

-- AI

5. Oct 12, 2004

### galois427

ok, this is what i got.

AX x AZ = (AB+BX) x (AD+DZ)
= (AB+BX) x AD + (AB+BX) x DZ
= AB x AD + BX x AD + AB x DZ + BX x DZ

AB x AD = (AX+XB) x (AZ+ZD)
= (AX+XB) x AZ + (AX+XB) x ZD
= AX x AZ + XB x AZ + AX x ZD + XB x ZD

and thats where i get stuck. how do you go about proving that those 4 vectors are equal to the other 4 vectors? are you suppose to put in arbitary numbers?

6. Oct 12, 2004

### robphy

You may be able to turn this geometric observation into a proof.

Extend ZDY and BXC and mark their intersection as M.
Imagine sliding AB to AX and DC to DM.
Parallelogram AXMD has the same area as ABCD.
Now slide XM to XY and AD to AZ.
Parallelogram AXYZ has the same area as AXMD.

7. Oct 13, 2004

### TenaliRaman

Ah indeed robphy!

galois,
to continue with the idea ...
what is BX x AD? (hint : definition of cross product)
what is AB x DZ + BX x DZ ??
(hint : again u can use the definition of cross product here ... )

-- AI

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