Bài 1:
Ta có AD là phân giác của góc A
=> \(\dfrac{AB}{AC}=\dfrac{DB}{DC}=\dfrac{15}{20}=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{AB}{AC}=\dfrac{3}{4}\Rightarrow\dfrac{AB^2}{AC^2}=\dfrac{9}{16}\)
\(\Rightarrow\dfrac{AB^2}{AC^2+AB^2}=\dfrac{9}{25}\)
\(\Rightarrow\dfrac{AB^2}{BC^2}=\dfrac{9}{25}\)
\(\Rightarrow AB=\sqrt{35^2.\dfrac{9}{25}}=21\)
ebsfbeddbaw dbfjefbeb fbyesnfiuhfsenfeshfkbnfnhfsbvjsnvni ưojf jfwnfieyuhfnrhghre fjeohfgvjsijujehg grhuihgebysuoghygwg ủedhngulhejwzpfohaf rudgohjguiehgeaoov gjrpfbr0-ehQAhnfwng jjtgt4tnfapwazjg0arp jer9gja9jmfejvewswfv gkfrdoegjriuigrgijreuhfo4wh4 euihfhehfjwdujwrifriwejfojesohfn 9we8ehfuewijfnagyuheofuwfeuhweucuidsuxvjfps9[ừhnqwln ecwe ewuijwiwqiwebfbwbufe]fghbyzelufnywegbf ewyufhewasuyfabwe;ăzhohyghflhw'flwefpawif3fj0[ư;fj[;kừbiwhfbqpfoebufiiobgfegfuyebwoghfyuweibfg iyfaygn98 ỳg07r m0rh 0mtvhn08yensh 9nw7r
