Q.In $\triangle ABC$, if $a+b+c=2s$, then prove that $\sin\left(\dfrac{A}{2}\right)=\sqrt{\dfrac{(s-b)(s-c)}{bc}}$, with usual notations.

Writing $a+b+c=2s$ (so $s$ is the semi-perimeter), the half-angle formulae express the sine, cosine and tangent of half of any angle of a triangle purely in terms of its three sides: $\sin\frac A2=\sqrt{\frac{(s-b)(s-c)}{bc}}$, $\cos\frac A2=\sqrt{\frac{s(s-a)}{bc}}$, and $\tan\frac A2=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ (with the exactly analogous formulae for $B$ and $C$). They are derived by substituting the Cosine Rule's expression for $\cos A$ into the double-angle identities $1-\cos A=2\sin^2\frac A2$ and $1+\cos A=2\cos^2\frac A2$, then factoring the resulting difference or sum of squares in terms of $s$. Heron's Formula, $A(\triangle ABC)= …