Prove that the ratio of the areas of the triangles $A'B'C'$ and $ABC$ is $2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

Question

If the bisectors of the angles of a triangle $ABC$ meet the opposite sides in $A',B',C'$,prove that the ratio of the areas of the triangles $A'B'C'$ and $ABC$ is $2\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}:\cos \frac{A-B}{2}\cos \frac{B-C}{2}\cos \frac{C-A}{2}$

Area of $ABC=$Area of $ABA'+$Area of $ACA'$
$=\frac{1}{2}(c+b)AA'\sin \frac{A}{2}=R(\sin C+\sin B)AA'\sin \frac{A}{2}$
$=R\times 2\sin \frac{B+C}{2} \cos \frac{B-C}{2}AA'\sin \frac{A}{2}$

How to find the area of $A'B'C'$ and get the desired result,help me please.Thanks in advance.

Answer 1

Let me try. One has $$\frac{B'A}{B'C} = \frac{BA}{BC} \Rightarrow \frac{B'A}{AC} = \frac{BA}{BA+BC},$$ $$\frac{C'A}{C'B} = \frac{CA}{CB} \Rightarrow \frac{C'A}{AB} = \frac{CA}{CA+CB}$$.

So, one has $$\frac{S_{AB'C'}}{S_{ABC}} = \frac{AB'.AC'}{AC.AC} = \frac{BA.CA}{(BA+BC)(CA+CB)}.$$

Similarly, one has

$$\frac{S_{A'BC'}}{S_{ABC}} = \frac{BA'.BC'}{BA.BC} = \frac{BA.BC}{(AB+AC)(BC+AC)}.$$

$$\frac{S_{A'B'C}}{S_{ABC}} = \frac{CA'.CB'}{CA.CB} = \frac{CA.CB}{(BC+AB)(CA+AB)}.$$

Thus, $$\frac{S_{A'B'C'}}{S_{ABC}} = 1 - \frac{S_{AB'C'}}{S_{ABC}} - \frac{S_{A'BC'}}{S_{ABC}} - \frac{S_{A'B'C}}{S_{ABC}} = \frac{(AB+BC)(BC+CA)(CA+AB) - AB.CA(AB+CA)- AB.BC(AB+BC)-CA.BC(BC+CA)}{(AB+BC)(BC+CA)(CA+AB)}$$

$$= \frac{2AB.BC.CA}{(AB+BC)(BC+CA)(CA+AB)}$$

$$ = \frac{2\sin A\sin B\sin C}{(\sin A + \sin B)(\sin B + \sin C)(\sin A + \sin C)}$$

$$ = 2\frac{\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}}{\cos \frac{A-B}{2}\cos \frac{B-C}{2}\cos \frac{C-A}{2}} .$$