Euler counting for a finite planar triangulation

Let $h$ be the number of hull corners, meaning the boundary sites that are not in the relative interior of any hull edge. Their interior angles sum to $(h-2)\pi$, because a convex $h$-gon is divided into $h-2$ triangles by diagonals from one corner. Each of the other $k-h$ boundary sites has straight angle $\pi$. Around each interior site the incident triangle angles sum to $2\pi$, while at each boundary site they sum to its hull angle. Counting all triangle angles gives $\pi T=2\pi (n-k)+(h-2)\pi +(k-h)\pi =(2n-k-2)\pi$, hence $T=2n-2-k$.

Count incidences between triangle sides and edges. Each triangle contributes three. An interior edge has one triangle on each of its two local sides; a hull edge has exactly one incident triangle inside the convex hull. Since the hull boundary has $k$ sites, it has $k$ subdivided boundary edges. Thus $3T=2(E-k)+k=2E-k$.

Substituting the value of $T$ from step 1.1 into step 1.2 gives $E=3n-3-k$. Consequently $n-E+T=n-(3n-3-k)+(2n-2-k)=1$, proving both identities.