Sửa đề: Chứng minh a+d=b+c
(a+b+c+d)(a-b-c+d)=(a-b-c+d)(a+b-c-d)
=>\(\left(a+d\right)^2-\left(b+c\right)^2=\left(a-c\right)^2-\left(b-d\right)^2\)
=>\(\left(a+d\right)^2-\left(a-c\right)^2=\left(b+c\right)^2-\left(b-d\right)^2\)
=>\(\left(a+d-a+c\right)\left(a+d+a-c\right)=\left(b+c-b+d\right)\left(b+c+b-d\right)\)
=>\(\left(d+c\right)\cdot\left(2a+d-c\right)=\left(c+d\right)\cdot\left(2b+c-d\right)\)
=>2a+d-c=2b+c-d
=>2a-2b=c-d-d+c=2c-2d
=>a-b=c-d
=>a+d=b+c
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