prove that for every parallelogram

Asked Dec 25 '19 at 14:27

Active Dec 25 '19 at 15:13

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Prove that for every parallelogram $$d_1^2+d_2^2=2(a^2+b^2)$$ My attempt is:

$d_1^2=a^2+b^2-2ab\cos\alpha$

$d_2^2=a^2+b^2-2ab\cos(180-\alpha)=a^2+b^2+2ab\cos\alpha$

It follows that: $$d_1^2+d_2^2=2(a^2+b^2)$$

but I don't know whether that is correct. Please help me.

edited Dec 25 '19 at 15:13

Bernard

asked Dec 25 '19 at 14:27

betmen 12

That's a fine solution. Was that your question? – Dec 25 '19 at 14:30
why the second angle is $180-\alpha$, or becouse the sum of tha angle in the base of paralelogram is 180 – betmen 12 Dec 25 '19 at 14:36
The adjacent internal angles in a parallelogram are supplementary. They are same-side interior angles if you consider their common side as a transversal cutting their opposite sides, which are parallel. – Dec 25 '19 at 14:38
thanky very much sir – betmen 12 Dec 25 '19 at 14:39

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