How to show that the set of extreme points of $\mathcal{P}_{1}(\mathcal{A})$ is precisely $\operatorname{Proj}(\mathcal{A})$. The problem is as follows:

Let $\mathcal{A}$ be a $C^{*}$-algebra. Note that
$\operatorname{Proj}(\mathcal{A})$ is contained in the set $\mathcal{P}_{1}(\mathcal{A}):=\left\{x \in \mathcal{A}^{+}:\|x\| \leq 1\right\}$, and that $\mathcal{P}_{1}(\mathcal{A})$ is closed and convex.
(i) Let $y \in \mathcal{A}^{+}$and let $p \in \operatorname{Proj}(\mathcal{A})$.
Suppose that $\left(1_{\mathcal{A}}-p\right)
y\left(1_{\mathcal{A}}-p\right)=0$. Show that $y=p y p$.
(ii) Let $x, y \in \mathcal{A}^{+}$be such that $x+y=0$. Show that $x=y=0$.
(iii) Show that the set of extreme points of
$\mathcal{P}_{1}(\mathcal{A})$ is equal to
$\operatorname{Proj}(\mathcal{A})$. [Hint: Use continuous functional calculus to show that non-projections are not extreme points. In the other direction, first prove that the unit is extreme, and then use parts (i) and (ii).]
(iv) Show that conv$\operatorname{Proj}(\mathcal{A})=\mathcal{P}_{1}(\mathcal{A})$, when $\mathcal{A}=M_{n}(\mathbb{C})$, for some $n \geq 2$.
(v) Find an example of a $C^{*}$-algebra $\mathcal{A}$ for which conv$\operatorname{Proj}(\mathcal{A})$ is not dense in $\mathcal{P}_{1}(A)$.

I got stuck at (iii). I proved that the unit is indeed extreme, but I don't know how to use parts (i) and (ii) to prove the conclusion for general elements $p \in \operatorname{Proj}(\mathcal{A})$.
Any help is appreciated.

Assume $p=(1-t)a+tb,$ where $0<t<1$ and $0 \le a,b \le I.$ Then $$0=(1-t)(I-p)a(I-p)+t(I-p)b(I-p).$$ Therefore $$(I-p)a(I-p)=(I-p)b(I-p)=0.$$ Observe  that $$ (I-p)a(I-p)= [a^{1/2}(I-p)]^*a^{1/2}(I-p).$$ Hence $ 
a^{1/2}(I-p)=0.$ We get $a(I-p)=0.$ Thus $a=ap$ and $a=(ap)^*=pa.$ This implies $a\le p$ as $p-a=p-ap=(I-a)p.$ Similarly $b\le p.$ We have $$0=(1-t)(p-a) + t(p-b) .$$ Therefore both summands vanish i.e. $p=a=b.$
