Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 \(\frac{AD}{\sin C}=2R\)
=>\(2R=\frac{a\sqrt7}{3}:\sin60=\frac{a\sqrt7}{3}:\frac{\sqrt3}{2}=\frac{a\sqrt7}{3}\cdot\frac{2}{\sqrt3}=\frac{2a\sqrt7}{3\sqrt3}\)
=>\(R=\frac{a\sqrt7}{3\sqrt3}\)
\(\frac{R}{r}=\frac{a\sqrt7}{3\sqrt3}:\frac{a\sqrt3}{5+\sqrt7}=\frac{\sqrt7\left(5+\sqrt7\right)}{3\sqrt3\cdot\sqrt3}=\frac{5\sqrt7+7}{9}\)
b: Đặt AB=AC=BC=a
\(\overrightarrow{DC}=2\cdot\overrightarrow{BD}\)
=>\(\overrightarrow{DC}=-2\cdot\overrightarrow{DB}\)
=>D nằm giữa B và C sao cho DC=2DB
DB+DC=BC
=>BC=2DB+BD=3BD
=>\(BD=\frac{a}{3}\)
=>\(CD=\frac23a\)
ΔABC đều
=>\(\hat{ABC}=\hat{ACB}=\hat{BAC}=60^0\)
Xét ΔACD có \(cosC=\frac{CA^2+CD^2-AD^2}{2\cdot CA\cdot CD}\)
=>\(a^2+\left(\frac23a\right)^2-AD^2=2\cdot a\cdot\frac23a\cdot cos60=\frac43a^2\cdot\frac12=\frac23a^2\)
=>\(AD^2=a^2+\frac49a^2-\frac23a^2=a^2-\frac29a^2=\frac79a^2\)
=>\(AD=\frac{a\sqrt7}{3}\)
Xét ΔCAD có \(\frac{AD}{\sin C}=2R\)
=>\(2R=\frac{a\sqrt7}{3}:\sin60=\frac{a\sqrt7}{3}:\frac{\sqrt3}{2}=\frac{a\sqrt7}{3}\cdot\frac{2}{\sqrt3}=\frac{2a\sqrt7}{3\sqrt3}\)
=>\(R=\frac{a\sqrt7}{3\sqrt3}\)
\(S_{CAD}=\frac12\cdot CA\cdot CD\cdot\sin C\)
\(=\frac12\cdot a\cdot\frac23a\cdot\sin60=\frac13\cdot a^2\cdot\frac{\sqrt3}{2}=\frac{a^2\sqrt3}{6}\)
\(p=\frac{CA+CD+AD}{2}\)
\(=\frac12\left(a+\frac23a+\frac{a\sqrt7}{3}\right)=\frac12\cdot a\cdot\left(\frac53+\frac{\sqrt7}{3}\right)=\frac{a}{2}\cdot\frac{5+\sqrt7}{3}=\frac{a\left(5+\sqrt7\right)}{6}\)
\(S=p\cdot r\)
=>\(r=\frac{a^2\sqrt3}{6}:\frac{a\left(5+\sqrt7\right)}{6}=\frac{a^2\sqrt3}{a\left(5+\sqrt7\right)}=\frac{a\cdot\sqrt3}{5+\sqrt7}\)
\(\frac{R}{r}=\frac{a\sqrt7}{3\sqrt3}:\frac{a\sqrt3}{5+\sqrt7}=\frac{\sqrt7}{3\sqrt3}\cdot\frac{5+\sqrt7}{\sqrt3}=\frac{5\sqrt7+7}{9}\)