Question

ABCD is a quadrilateral. E, F are the midpoints of AC and BD respectively. Prove that AB^2 + BC^2 + CD^2 +DA^2=AC^2+BD^2+4EF^2

Answer 1

@SolomonZelman @jim

Answer 2

@jim_thompson5910 my bad

Answer 3

@zzr0ck3r

Answer 4

never taken a Geometry class.

Answer 5

oh by the way, do all this using only algebra

Answer 6

and fact about geometry I am sure....

Answer 7

@iambatman @iGreen @TuringTest

Answer 8

@mathmate

Answer 9

@wio

Answer 10

@Compassionate

Answer 11

@surjithayer you can only use algebra

Answer 12

Start with drawing a diagram of the quadrilateral ABCD, with the mid points of AC and BD named E and F respectively.
Without loss of generality (WLOG), we can place A at (0,0), and B on the x-axis.
We end up with the following coordinates:
\(A(0,0)\)
\(B(b,0)\)
\(C(c,r)\)
\(D(d,s)\)
\(E(\frac{c}{2}, \frac{r}{2})\)
\(F(\frac{d+b}{2}, \frac{s}{2})\)

Answer 13

I will do the first part, using distance between two points:
\(AB^2+BC^2+CD^2+DA^2\)
\(=(b^2)+((c-b)^2+r^2)+((d-c)^2+(s-r)^2)+(d^2+s^2)\)
\(=2s^2-2rs+2r^2+2d^2-2cd+2c^2-2bc+2b^2\)

You can complete the same exercise for the right-hand side:
\(AC^2+BD^2+4EF^2~=~...\)
and show that the two parts are equal.

Answer 14

E is the mid-point of AC, so E((0+c)/2, (0+r)/2)=(c/2,r/2)
F is the mid-point of BD, so F((b+d)/2, (0+s)/2)=((b+d)/2, s/2)

Answer 15

THe algebra might be a little messy when it comes to the length of \(EF^2\), but it is rewarding at the end of the calculations.

Answer 16

Yep got it thanks so much

Answer 17

You're welcome! :)