Why is the sum of squares of the diagonals of a parallelogram equal to the sum of squares of its four sides? I don't know what the cosine theorem is and I didn't learn it!

a^2+b^2-2ab cos A=l1^2
a^2+d^2-2ad cos (180-A)=l2^2
c^2+b^2-2cb cos (180-A)=l2^2
c^2+d^2-2cd cos A=l1^2
The four formulas add up
Because cos a + cos (180-a) = 0, we can get the answer

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