Give a proof of the theorem of the relationship between the diagonal and the sides of a parallelogram (the sum of the squares of the two diagonals of a parallelogram is equal to the sum of the squares of its four sides)

Using Cosine Theorem

Verification: The sum of squares of two diagonals of a parallelogram is equal to the sum of squares of four sides.

It is known that in parallelogram ABCD, AC, BD are their two diagonal,
Verification: AC2+BD2=AB2+BC2+CD2+AD2
The results showed that AE⊥BC was at point E, DF⊥BC was at point F,
Then ∠AEB=∠DFC=90°.
Quadrilateral ABCD is a parallelogram,
AB=DC, AB//CD,
ABE=∠DCF,
ABE DCF,
AE=DF, BE=CF.
In Rt△ACE and Rt△BDF, from Pythagorean theorem,
AC2= AE2+EC2= AE2+(BC-BE)2,
BD2= DF2+BF2= DF2+(BC+CF)2= AE2+(BC+BE)2,
AC2+BD2=2AE2+2BC2+2BE2=2(AE2+BE2)+2BC2.
AE2+BE2=AB2,
I.e. AC2+BD2=2(AB2+BC2).
AB=CD, AD=BC,
AC2+BD2=AB2+BC2+CD2+AD2

It is proved by vector that the sum of squares of two diagonals of a parallel four-sided line is equal to the sum of squares of four sides It is proved by vector that the sum of squares of two diagonals of a parallel four-sided row is equal to the sum of squares of four sides

Set parallelogram ABCD
Then AC^2+BD^2=(AB+BC)^2+(BA+AD)^2=AB^2+BC^2+2AB*BC*cos (π-B)+BA^2+AD^2+2BA*AD*cos (π-A)=AB^2+BC^2+CD^2+DA^2-cosB2AB*BC+2BA*AD*cosB
=AB^2+BC^2+CD^2+DA^2
Certificate obtained

Prove that the sum of squares of four sides of a parallelogram is equal to the sum of two diagonals Suppose you only have junior high school knowledge, only geometric proof, thank you. Prove that the sum of squares of four sides of a parallelogram equals the sum of two diagonals Suppose you only have junior high school knowledge, only geometric proof, thank you.

Certification: as shown in the figure

Make the height of BC edge over A and D, and the vertical foot is E and F respectively.

△ABE DCF is easily known

BE=CF, AE=DF

By Pythagorean Theorem

BD2=BF2+DF2

BD2=(BC+CF)2+DF2

=BC2+2*BC*CF+CF2+D F2

AC2=AE2+CE2

= AE2+(BC-BE)2

=AE2+BC2-2*BC*BE+BE2

Therefore, BD2+AC2=(BC2+2*BC*CF+CF2+DF2)+(AE2+BC2-2*BC*BE+BE2)

=2*BC2+2(CF2+DF2)

=2*BC2+2*CD2

= BC2+AD2+AB2+CD2

I.e.

BD2+AC2=BC2+AD2+AB2+CD2

How to prove by analytic method that the sum of squares of the sides of a parallelogram is equal to the sum of squares of two diagonals

Vector AB = vector e1
Vector AD = vector e2
Vector AC = vector e1+ vector e2
Vector BD = vector e1- vector e2
AC+BD=(vector e1+vector e2)+(vector e1-vector e2)=2(e1+e2)=AB+BC+CD+DA

Why if aOA+bOB+cOC=0(OA,OB,OC,0 are all vectors), then O is the heart of the triangle ABC? After the equal sign is the zero vector How the hell did you get it out? Why if aOA+bOB+cOC=0(OA,OB,OC,0 are all vectors), then O is the heart of the triangle ABC? After the equal sign is the zero vector How the hell did it come out?

First find out the inner position, and substitute it into it. Then there is only one point O to establish the above equation, so the O satisfying the equation is the inner.
As for the inner solution, see the link below, whether pure vector or coordinate system, the method is the same.

Give a proof of the theorem of the relationship between the diagonal and the sides of a parallelogram (the sum of the squares of the two diagonals of a parallelogram is equal to the sum of the squares of its four sides)