In his Geometria Dominicana blog, Emmanuel José García has been championing a pair of half-angle identities
in a triangle. I thought I’d contribute to the cause by offering trigonographic proofs.

$$\begin{align}|CB’||CD|=|CQ||CQ’| &\quad\to\quad 2a\cos\frac12C\cdot2b\cos\frac12C=(a+b+c)(a+b-c) \\[8pt] &\quad\to\quad \cos^2\frac12C=\frac{s(s-c)}{ab} \qquad\qquad \left(\;s:=\frac12(a+b+c)\;\right) \end{align}$$

$$\begin{align}|PB||PA’|=|PQ||PQ’| &\quad\to\quad 2a\sin\frac12C\cdot2b\sin\frac12C=(-a+b+c)(a-b+c) \\[8pt] &\quad\to\quad \sin^2\frac12C=\frac{(s-a)(s-b)}{ab}\end{align}$$