Rectangle ABCD contains a point X such that AX=1, BX=7, and CX=8. Find DX, then prove DX is the same value no matter what size and shape rectangle you have that fulfills these requirements.

I solved for DX, but I don't know how to do the second part.

AngleObtuse Mar 1, 2021

#1**+1 **

I don't know how you solved it, but the way I solved it directly proves that DX is the same value.

So here is what I did:

(the green, blue, red, and purple lines all intersect perpendicularly)

Let the green side be g, the blue side be b, the red side be r, and the purple side be p.

Then

\(g^2+p^2=1\\ b^2+g^2=49\\ r^2+b^2=64\\\)

subtract the second equation from the first equation, then add the difference to the third equation to get

\(DX=\sqrt{1-49+64}\\DX=4\)

For the second part of the question, suppose DX is not 4. The algebra above shows that it is 4, so it's a contradiction.

(like mentioned before the way I solved it makes the proof self explanatory)

textot Mar 1, 2021