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Sa se arate ca un triunghi \( ABC \) este dreptunghic \( \Longleftrightarrow \ \cos\frac A2\cos\frac B2\cos\frac C2-\sin\frac A2\sin\frac B2\sin\frac C2=\frac 12 \) .

Folosim identitatile: \( \left\|\ \begin{array}{cc} \cos\frac A2\cos\frac B2\cos\frac C2 & = & \frac{p}{4R}\\\\\\\\\
\sin\frac A2\sin\frac B2\sin\frac C2 & = & \frac{r}{4R}\ \end{array}\right\| \) .

Astfel egalitatea devine : \( p=2R+r\ \Longleftrightarrow\ p^{2}=4R^{2}+4Rr+r^{2}\ \Longleftrightarrow\ 2p^{2}-8Rr-2r^{2}=8R^{2} \)

\( \Longleftrightarrow\ a^{2}+b^{2}+c^{2}=8R^{2}\ \Longleftrightarrow\ \sin^{2}A+\sin^{2}B+\sin^{2}C=2\ \Longleftrightarrow\ \cos 2A+\cos 2B+\cos 2C+1=0 \)

\( \Longleftrightarrow\ 2\cos(A-B)\cos (A+B)+2cos^{2}C=0\ \Longleftrightarrow\ \cos C=0 \) sau \( \cos C=\cos(A-B) \)

\( \Longleftrightarrow\ C=90^{\circ} \) sau \( A=B+C \) sau \( B=A+C \) \( \Longleftrightarrow\ 90^{\circ}\in\{A,\ B,\ C} \) .

Se arata usor prin relatiile Viete ca ecuatia \( f(x)=p\cdot x^3-(4R+r)\cdot x^2+p\cdot x-r=0 \) admite

radacinile \( \left\{\tan\frac A2\ ,\ \tan\frac B2\ ,\ \tan\frac C2\right\} \) . Se observa ca \( f(1)=2[p-(2R+r)] \) . Asadar,

\( p=2R+r\ \Longleftrightarrow\ f(1)=0\ \Longleftrightarrow\ 1\in \left\{\tan\frac A2\ ,\ \tan\frac B2\ ,\ \tan\frac C2\right\}\ \Longleftrightarrow\ 90^{\circ}\in\{A,B,C\} \) .