Given $a_i\in\mathbb{R}^n,b_i\in\mathbb{R},1\leq i\leq p$, define a polyhedron $P=\{x\in\mathbb{R}^n|\langle a_i,x\rangle\leq b_i,1\leq i\leq p\}$ and $I=\{i|\langle a_i,x\rangle=b_i,\forall x\in P\}$, define $A=\{x\in\mathbb{R}^n|\langle a_i,x\rangle\leq b_i,\forall i\in I\}$, prove that $\text{aff}(P)=A$ (Here aff means the affine hull).
It is esay to prove that $\text{aff}(P)\subset A$ because $P\subset A$ and $\forall x_1,x_2\in P,\forall\lambda\in\mathbb{R},\lambda x_1+(1-\lambda)x_2\in A$. For the other side, we can prove $I=\{i|\langle a_i,x\rangle=b_i,\forall x\in \text{aff}(P)\}$, but how to prove that any $x\in A-P$ can be written as the affine combination of elements in $P$?
Notation: $H(a,b):=\{x\in\mathbb{R}^n|\langle a,x\rangle= b\}$, $K(a,b):=\{x\in\mathbb{R}^n|\langle a,x\rangle\le b\}$, $riC$-the relative interior of $C$. $dim(P)$-the dimension of the affine hull of $P$.
Lemma: Let $S$ be convex, and $H$, a half space. If $S\cap riH\ne \phi$, then $dim(S)=dim(S\cap H)$.
Proof: Let $dim(S)=k$. Then, there is a $k+1$-affine independent family, $\lbrace y_1,\cdots,y_{k+1}\rbrace$, in $S$ with atleast one member in $S\cap riH$. To see this, choose an $k+1$-affine independent family in $S-riH$. Then, since $S\cap riH$ is non-empty, there is an $x\in S\cap riH$ and it can be written as $$x=c_1y_1+\cdots+c_{k+1}y_{k+1}\tag 1$$ for $\sum c_i=1$.
Let $c_1\ne 0$, WLOG. Then, $x,y_2,\cdots,y_{k+1}$ is affinely independent. Otherwise, if $d_1x+d_2y_2+\cdots+d_{k+1}y_{k+1}=0$ for some $\sum d_i=0$. But this would mean that $x_1,\cdots,x_{k+1}$ are affinely dependent, substituting from $(1)$.
Now, choose an affine independent family $x,y_2,\cdots,y_{k+1}$ such that $x\in S\cap ri(H)$. Clearly, there is a $\lambda$ such that the segment $[x,\lambda x+(1-\lambda)y_i]$ $\forall 2 \le i\le k+1$ is completely in $S\cap riH$. The points $x,\lambda x+(1-\lambda)y_2,\cdots,\lambda x+(1-\lambda)y_{k+1}$ is an affine independent family in $S\cap H$ of size $k+1$. This finishes the proof of the lemma.
Using the lemma, we show that $$\text{aff}P=\cap_{i\in I} H(a_i,b_i).$$
Why? Because $(\cap_{i\notin I} K(a_i,b_i))\cap (\cap_{i\in I}H(a_i,b_i))=P$ and repeatedly applying the lemma (because $P\cap riK(a_i,b_i)\ne \phi \forall i\notin I$) will give $$dim(P)=dim((\cap_{i\notin I} K(a_i,b_i))\cap (\cap_{i\in I}H(a_i,b_i)))=dim(\cap_{i\in I}H(a_i,b_i)).$$
Since $\text{aff}P\subset \cap_{i\in I}H(a_i,b_i)$ and both are affine, the equality of dimensions means that they are equal.
We finish off by showing that $A=\cap_{i\in I} H(a_i,b_i)$.
If not and there is an element $x\in A-\cap_{i\in I} H(a_i,b_i)$, then $dim(A)>dim(\cap_{i\in I} H(a_i,b_i))$, as $\cap_{i\in I} H(a_i,b_i)$ is affine. Using the lemma once again will give that $$dim(P)=dim(\text{aff}P)=dim((\cap_{i\notin I} K(a_i,b_i))\cap (\cap_{i\in I}K(a_i,b_i)))$$ $$=dim(\cap_{i\in I}K(a_i,b_i))>dim(\cap_{i\in I}H(a_i,b_i))=dim(\text{aff}P)=dim(P).$$