omriya200 wrote: $AD$ Is the angle bisector Of $\angle BAC$ Where $D$ lies on the The circumcircle of $\triangle ABC$. Show that $2AD>AB+AC$ We need to prove that $$4R\sin\left(\beta+\frac{\alpha}{2}\right)>b+c$$or $$\frac{abc}{S}\left(\frac{2S}{ac}\cdot\sqrt{\frac{1+\frac{b^2+c^2-a^2}{2bc}}{2}}+\frac{a^2+c^2-b^2}{2ac}\cdot\sqrt{\frac{1-\frac{b^2+c^2-a^2}{2bc}}{2}}\right)>b+c$$or $$\sqrt{\frac{b(a+b+c)(b+c-a)}{c}}+\frac{a^2+c^2-b^2}{4S}\cdot\sqrt{\frac{b(a+b-c)(a+c-b)}{c}}>b+c$$or $$\sqrt{\frac{b(a+b+c)(b+c-a)}{c}}+(a^2+c^2-b^2)\sqrt{\frac{b}{c(a+b+c)(b+c-a)}}>b+c$$or $$\sqrt{b}(b^2+2bc+c^2-a^2+a^2+c^2-b^2)>(b+c)\sqrt{c(b^2+2bc+c^2-a^2)}$$or $$2\sqrt{bc}>\sqrt{b^2+c^2+2bc-a^2}$$or $$(a+b-c)(a+c-b)>0.$$Done!

omriya200 wrote: $AD$ Is the angle bisector Of $\angle BAC$ Where $D$ lies on the The circumcircle of $\triangle ABC$. Show that $2AD>AB+AC$

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Dear Mathlinkers, just Ptolemee's theorem and majoration... Sincerely Jean-Louis

jayme, see the mihaig's solution. I think it's the easiest way.

Dear, thank for your observation... In my point of view, I prefer to work without angle... Sincerely Jean-Louis

Let $I$ and $I_A$ be the incenter and A-excircle of $\triangle{ABC}$ $$2AD=AI_A+AI>\frac{a+b+c}{2}+\frac{b+c-a}{2}=b+c$$

arqady wrote: jayme, see the mihaig's solution. I think it's the easiest way. Thank you,professor Rozenberg. But if the maestro doesn't like the angles,neither of us is able to pass the grade.

A purely synthetic solution is possible! Best regards, sunken rock