```Theorem 2:

If A, B, C are collinear, then real numbers x, y, z not all zero such that (bidirectional)
x+y+z=0 and xA + yB + zC = 0

Dot (Inner) Product:

A . B is equal to |A| |B| cos(angle between). Which is essentially the product of the projection of one vector onto the other.

Exterior Product (wedge product):
u ^ u = 0
u ^ v + v ^ u = 0
Length: |u| * |v| * sin (angle)
Geometrically: it’s the area of the quadrilateral created by the two vectors.
Algebraically, determinant with z column zero: u1*v2 - u2*v1.

Law of Cosines:

Given an angle, theta, between b and a, c^2 = a^2 + b^2 - 2abcos(theta).

REVISIT TRIPLE PRODUCT OF 3 3D VECTORS.
Also written wedge (^). Is equal to twice the area of the triangle formed by the two vectors.
A ^ B ^ C = 1/2 the area of the triangle ABC in the Z=1 plane.

if u,v,w in R^2, then u ^ v ^ w = 0.
A,B,C collinear (3 points in the plane) <==> A ^ B + B^C  + C^A = 0.
Need to get the whole collinearity and determinant down.

Barycentric Coordinates:

A, B, C three non-collinear points, any vector P may be expressed as P = xA + yB + zC, where x + y + z = 1.

Altitudes of triangles are concurrent proof: Draw the triangle, draw two of the altitudes and the point that they intersect at, and then you’ll say “let’s show that the line that goes through this intersection from the remaining vertex is actually an altitude”. So you just need to show that some dot product is zero. Should be algebra.

Centroid proof: So like, for the centroid of a triangle being the average of the three sides, you connect the edges to the midpoint of their opposite side, and you notice that the intersection seems to divide these lines in a ration of 2:1, so you write the midpoint as two of the sides and then you’re tasked with incorporating the last side, so you do so by considering this point on the line with a ratio of 2:1, and then you see that the algebra simplifies to sum of all the sides over three.

Medians of a triangle are concurrent proof: Really, you just observe the median for one of the sides, and you consider the point along the anchor that is 2:1 farther from the vertex it drops from. Then you’ll see that because it bisects the opposite side, you can write it as sum of halves of the other two sides. It’s algebra from there. This is the triangle’s centroid.

Black hole problem: two lines and a point, all intersecting at a point, describe that point of intersection by using Disargues. Short version: create two triangles with these lines, show that they’re perspective with respect to a line and therefore they are perspective with respect to a point. Essentially what you do is create one full triangle with your lines and then another side of a triangle. You show that these two have an axis of perspective, MAY WANT TO REVISIT.

REVISIT THEOREMS FROM CLASS I MISSED.

feuerbach’s circle theorem: Just take a brief look.

Menelaus Proof: Use Theorem 2.

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