It's best to read about Caratheodory's theorem, to know we're trying to prove.
Caratheodory's theorem on convex sets is essentially equivalent to Radon's theorem; we'll use that theorem in our proof.

Suppose x = ∑_{k=1}^{m}x_{k}x_{k} is a convex combination of m>d+1 points. We need to show that x is a convex combination of m-1 of these points (this is the equivalent formulation given).

By Radon's theorem (and since m>d+1 is large enough!), we can divide our m points into 2 nonempty sets of points (WLOG x_{1},...,x_{l} and x_{l+1},...,x_{m}) whose convex hulls share a point:

∑_{i=1}^{l}a_{i}x_{i} = ∑_{i=l+1}^{m}b_{i}x_{i}

a_{i} >= 0, b_{j} >= 0, ∑a_{i} = ∑b_{j} = 1

Furthermore, one of the b's, WLOG b

_{m}, satisfies ∀k:c

_{k}/b

_{k} >= c

_{m}/b

_{m} (every finite set has a

minimum!).

Isolating x_{m}, we have:

x_{m} = (∑_{i=1}^{l}a_{i}x_{i} - ∑_{i=l+1}^{m-1}b_{i}x_{i}) / b_{m}

Substituting for x, we may write x as a

linear combination of the points x

_{1},...,x

_{m-1} (this in itself is no surprise; it's true for any linear combination of >d points in **R**^{d}! The point is we'll show it's a convex combination...):

x = ∑_{i=1}^{l}(c_{i}+a_{i}c_{m}/b_{m})x_{i} +
∑_{i=l+1}^{m-1}(c_{i}-b_{i}c_{m}/b_{m})x_{i}

Checking, we find the sum of the

coefficients is 1

(this is not surprising), so it's an

affine combination.

It remains to prove that all coefficients are nonnegative. Those involving a_{i} are just sums of positive numbers, so there's no problem. And for each l+1<=i<m, we have c_{i}>=b_{i}c_{m}/b_{m}) due to how we arranged the coefficients.

So x is a convex combination of m-1 points. We may continued to eliminate points from the convex combination until only d+1 are left; at that point, Radon's theorem could fail, so no further reduction is possible.