a: ΔABC vuông cân tại A
=>\(\widehat{ABC}=45^0\)
ΔABC vuông cân tại A
=>AB=AC=a và \(BC=a\cdot\sqrt{2}\)
=>\(BM=\dfrac{2}{3}\cdot BC=\dfrac{2}{3}\cdot a\sqrt{2}=\dfrac{2a\sqrt{2}}{3}\)
Xét ΔBAM có \(cosB=\dfrac{BA^2+BM^2-AM^2}{2\cdot BA\cdot BM}\)
=>\(\dfrac{a^2+\dfrac{8a^2}{9}-AM^2}{2\cdot a\cdot a\sqrt{2}\cdot\dfrac{2}{3}}=cos45=\dfrac{\sqrt{2}}{2}\)
=>\(\dfrac{17}{9}a^2-AM^2=\dfrac{\sqrt{2}}{2}\cdot2\sqrt{2}\cdot\dfrac{2}{3}\cdot a^2=\dfrac{4}{3}a^2\)
=>\(AM^2=\dfrac{17}{9}a^2-\dfrac{4}{3}a^2=\dfrac{5}{9}a^2\)
=>\(AM=\dfrac{a\sqrt{5}}{3}\)
b: \(cos\left(B+C\right)=-\dfrac{\sqrt{2}}{2}\)
=>\(\widehat{B}+\widehat{C}=135^0\)
=>\(\widehat{A}=180^0-135^0=45^0\)
Xét ΔABC có \(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
=>\(\dfrac{4+8-BC^2}{2\cdot2\cdot2\sqrt{2}}=cos45=\dfrac{\sqrt{2}}{2}\)
=>\(12-BC^2=8\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}=4\cdot2=8\)
=>\(BC^2=4\)
=>BC=2
