Đặt BC=a
ΔABC đều
=>AB=BC=AC=a
Ta có: DC+DB=BC
=>2DB+BD=BC
=>BC=3BD
=>\(DB=\frac{BC}{3}=\frac{a}{3}\)
=>\(DC=2\cdot DB=\frac{2a}{3}\)
Xét ΔCDA có \(cosC=\frac{CA^2+CD^2-AD^2}{2\cdot CA\cdot CD}\)
=>\(\frac{a^2+\left(\frac{2a}{3}\right)^2-AD^2}{2\cdot a\cdot\frac23a}=cos60=\frac12\)
=>\(a^2+\frac49a^2-AD^2=\frac12\cdot\frac43a^2=\frac23a^2\)
=>\(AD^2=\frac{13}{9}a^2-\frac23a^2=\frac{13}{9}a^2-\frac69a^2=\frac79a^2\)
=>\(AD=\frac{a\sqrt7}{3}\)
Nửa chu vi của tam giác CDA là:
\(p=\frac12\left(AC+CD+AD\right)=\frac12\left(a+\frac{2a}{3}+\frac{a\sqrt7}{3}\right)=\frac12\cdot\frac{5a+a\sqrt7}{3}=\frac{5a+a\sqrt7}{6}\)
Diện tích tam giác CDA là:
\(S_{CDA}=\frac12\cdot CA\cdot CD\cdot\sin C=\frac12\cdot a\cdot\frac23a\cdot\sin60\)
\(=\frac13a^2\cdot\frac{\sqrt3}{2}=\frac{a^2\sqrt3}{6}\)
Bán kính đường tròn nội tiếp ΔCDA là:
\(r=\frac{S_{CDA}}{p_{DAC}}=\frac{a^2\sqrt3}{6}:\frac{5a+a\sqrt7}{6}=\frac{a^2\sqrt3}{5a+a\sqrt7}=\frac{a\sqrt3}{5+\sqrt7}\)
Xét ΔCDA có \(\frac{AD}{\sin C}=2R\)
=>\(2R=\frac{a\sqrt7}{3}:\sin60=\frac{a\sqrt7}{3}:\frac{\sqrt3}{2}=\frac{a\sqrt7}{3}\cdot\frac{2}{\sqrt3}=\frac{2a\sqrt7}{3\sqrt3}\)
=>\(R=\frac{a\sqrt7}{3\sqrt3}\)
\(\frac{R}{r}=\frac{a\sqrt7}{3\sqrt3}:\frac{a\sqrt3}{5+\sqrt7}=\frac{\sqrt7\left(5+\sqrt7\right)}{3\sqrt3\cdot\sqrt3}=\frac{5\sqrt7+7}{9}\)
