I need to prove parallelogram law using the law of cosines.
2AB^2 + 2BC^2 = AC^2 + BD^2
Please, help me to do this.
Let
θ = angle DAB
Ø = angle ABC
By the law of cosines,
BD^2 = AD^2 + AB^2 - 2 AD AB cosθ
AC^2 = BC^2 + AB^2 - 2 BC AB cosØ
Now, since AD = BC,
BD^2 + AC^2 = 2AD^2 + 2AC^2 - 2 AD AB (cosθ + cosØ)
Since θ+Ø = 180°
cosθ = -cosØ
'nuff said.
Oops. In the long line, on the right side, AC should read AB
ehh, still don't get it.
where do you get stuck? I'll try to clarify. Draw a diagram to follow along.
To prove the parallelogram law using the law of cosines, first, let's construct a parallelogram ABCD with the sides AB, BC, CD, and AD.
We know that in a parallelogram, opposite sides are equal in length. Let's call the diagonals AC and BD intersecting at point O.
Now, consider triangle AOB. According to the law of cosines, we have:
AB^2 = OA^2 + OB^2 - 2(OA)(OB)cosθ1
Similarly, for triangle BOC:
BC^2 = OB^2 + OC^2 - 2(OB)(OC)cosθ2
Since OA = OC and OB = OD (opposite sides of the parallelogram are equal), we can rewrite the above equations as:
AB^2 = OA^2 + OB^2 - 2(OA)(OB)cosθ1
BC^2 = OB^2 + OD^2 - 2(OB)(OD)cosθ2
Now, let's add these two equations together:
AB^2 + BC^2 = OA^2 + OB^2 + OB^2 + OD^2 - 2(OA)(OB)cosθ1 - 2(OB)(OD)cosθ2
Simplifying further:
2AB^2 + 2BC^2 = 2(OA^2 + OB^2) + 2(OB^2 + OD^2) - 2(OA)(OB)cosθ1 - 2(OB)(OD)cosθ2
Notice that 2(OA^2 + OB^2) is equal to AC^2 (by the Pythagorean theorem) and 2(OB^2 + OD^2) is equal to BD^2 (also by the Pythagorean theorem).
Therefore, we have:
2AB^2 + 2BC^2 = AC^2 + BD^2 - 2(OA)(OB)cosθ1 - 2(OB)(OD)cosθ2
Now, since OA = OC and OB = OD, we can rewrite the equation as:
2AB^2 + 2BC^2 = AC^2 + BD^2 - 2(OA)(OB)(cosθ1 + cosθ2)
Since cosθ1 + cosθ2 = 0 (the diagonals of a parallelogram bisect each other at right angles), the term -2(OA)(OB)(cosθ1 + cosθ2) becomes 0.
Therefore, we finally have:
2AB^2 + 2BC^2 = AC^2 + BD^2
This proves the parallelogram law using the law of cosines.