Xét ΔADB có
\(cosA=\dfrac{AB^2+AD^2-DB^2}{2\cdot AB\cdot AD}\)
=>\(\dfrac{a^2+9a^2-DB^2}{2\cdot a\cdot3a}=\dfrac{1}{2}\)
=>\(10a^2-DB^2=3a^2\)
=>\(DB=a\sqrt{7}\)
Xét ΔABD có
\(cosABD=\dfrac{BA^2+BD^2-AD^2}{2\cdot BA\cdot BD}\)
\(=\dfrac{9a^2+7a^2-a^2}{2\cdot3a\cdot a\sqrt{7}}=\dfrac{15a^2}{6a^2\cdot\sqrt{7}}=\dfrac{15}{6\sqrt{7}}=\dfrac{5}{2\sqrt{7}}\)
=>\(cosCDB=\dfrac{5}{2\sqrt{7}}\)(do \(\widehat{ABD}=\widehat{CDB}\) vì AB//CD)
Xét ΔCDB có \(cosCDB=\dfrac{DB^2+DC^2-BC^2}{2\cdot DB\cdot DC}\)
=>\(\dfrac{5}{2\sqrt{7}}=\dfrac{7a^2+a^2-BC^2}{2\cdot a\sqrt{7}\cdot a}\)
=>\(\dfrac{8a^2-BC^2}{2a^2\sqrt{7}}=\dfrac{5}{2\sqrt{7}}\)
=>\(\dfrac{8a^2-BC^2}{a^2}=5\)
=>\(8a^2-BC^2=5a^2\)
=>\(BC^2=3a^2\)
=>\(BC=a\sqrt{3}\)
