Analytic Vector Proofs

[mitts_and_hats]

Hello, My techer left the class with a whole bunch of questions because he is gone for a week, and i am stuck on some of them. can some people help me with showing me how to do them?

1. Prove that the midpoint of the hypotenuse of a right-angled triangle is equidistant from each of the triangles verticies.

2. Prove the Appolonius Theorem: Double the sum of the squares of lengths of two sides of a triangle is equal to the sum of the square of the length of the third side and four times the length of the median subtended by the first two sides.

3. Please prove that the diagonals of a parallelogram bisect each other

4. Please Prove:If the Diagonals of parallelogram ABCD are perpendicular to one another, the ABCD is a rhombus

 

[stapel]

1) I'd start with the right triangle with the midpoint marked. Draw the perpendicular from the midpoint to the base of the triangle, cutting off a new right triangle. Then use similar triangles to start coming to some conclusions.

2) Sorry; I couldn't follow the statement of this exercise. (I should really be in bed by this time.) But if a theorem is famous enough to have a name, it's likely to have a proof somewhere online. Try Google.

3) Use vertical angle and alternate angles (the "lines crossing parallel lines" stuff) to show interior triangles to be similar, and then congruent. Make conclusions from this.

4) I'd go with similarity again.

Eliz.

 

[soroban]

Hello, mitts_and_hats!

#4 doesn't really need vectors . . .

. Prove: If the diagonals of parallelogram ABCD are perpendicular, then ABCD is a rhombus

Let \(\displaystyle O\) be the intersection of diagonals \(\displaystyle AC\) and \(\displaystyle BD.\)

In right triangle \(\displaystyle AOB:\;\;AO^2\,+\,OB^2\;=\;AB^2\;\;\Rightarrow\;\;AO^2\:=\:AB^2\,-\,OB^2\)

In right triangle \(\displaystyle AOD:\;\;AO^2\,+\,OD^2\:=\:AD^2\;\;\Rightarrow\;\;AO^2\:=\:AD^2\,-\,OD^2\)

. . Hence: .\(\displaystyle AB^2\,-\,OB^2\:=\:AD^2\,-\,OD^2\;\;\Rightarrow\;\;AB^2\:=\:AD^2\,+\,(OB^2\,-\,OD^2)\)


But \(\displaystyle OB\,=\, OD.\) . Diagonals of a parallelogram bisect each other (as proved in #3).

. . Therefore: .\(\displaystyle AB^2\,=\,AD^2\;\;\Rightarrow\;\;AB\,=\,AD.\) . \(\displaystyle ABCD\) is a rhombus.