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steven.wei

一道几何题。矩形abcd里面随便一点e,ae=x be=y ce=z de=w,求w^2=?? 用x^2, y^2, z^2表示的式子 w^2=x^2+z^2-y^2
[确认] old jj
不明白请指点

tongxun

b  _______________c
   | e-------------|o
a |_|h____________| d

I cannot draft it precisely and concisely.

de=w, de^2=w^2
w^=do^2 +e0^2
do^2=x^2 -ah^2
eo^2=z^2-co^
ah^2+co^2=y^2
Therefore, w^2=x^2 +z^2 -y^2