a) Ta có: $Ax//BC \, (gt)$
mà $BC\perp By \, (gt)$
nên $Ax\perp By$
$\Rightarrow \widehat{BDA} = 90^o$
Xét $ΔABC$ và $ΔDAB$ có:
$\widehat{BAC} = \widehat{BDA} = 90^o$
$\widehat{DAB} = \widehat{ABC}$ (so le trong)
Do đó $ΔABC\sim ΔDAB \, (g.g)$
b) Áp dụng định lý Pytago, ta được:
$BC^2 = AB^2 + AC^2 = 15^2 + 20^2 = 625$
$\Rightarrow BC = \sqrt{625} = 25 \, cm$
Ta có: $ΔABC\sim ΔDAB$ (câu a)
$\Rightarrow \dfrac{AB}{AD} = \dfrac{BC}{AB}$
$\Rightarrow AD = \dfrac{AB^2}{BC} = \dfrac{15^2}{25} = 9 \, cm$
Tương tự, ta có: $ΔABC\sim ΔDAB$
$\Rightarrow \dfrac{AC}{DB} = \dfrac{BC}{AB}$
$\Rightarrow BD = \dfrac{AC.AB}{BC} = \dfrac{15.20}{25} = 12 \, cm$
c) Ta có: $Ax//BC \, (gt)$
$\Rightarrow ΔDIA \sim ΔCIB$ (Theo định lý Thales)
$\Rightarrow \dfrac{S_{DIA}}{S_{BIC}} = \left(\dfrac{DA}{BC}\right)^2 = \dfrac{81}{625}$
$\Rightarrow S_{DIA} = \dfrac{81}{625}S_{BIC}$
Ta lại có:
$S_{BDC} - S_{BDA} = S_{BIC} + S_{DIA} = S_{BIC} +\dfrac{81}{625}S_{BIC} = \dfrac{706}{625}S_{BIC}$
Ta được:
$S_{BDC} - S_{BDA} = \dfrac{1}{2}(BC.DB - AD.DB) = \dfrac{1}{2}(25.12 - 9.12) = 96 \, cm^2$
$\Rightarrow \dfrac{706}{625}S_{BIC} = 96$
$\Rightarrow S_{BIC} = \dfrac{30000}{353} \approx 84,99 \, cm^2$
