Question

ý c) cần gấp

Answer 1

c: \(BD\cdot BK=BH\cdot BC\)

=>\(\dfrac{BD}{BC}=\dfrac{BH}{BK}\)

Xét ΔBDH và ΔBCK có

\(\dfrac{BD}{BC}=\dfrac{BH}{BK}\)

\(\widehat{DBH}\) chung

Do đó: ΔBDH~ΔBCK

=>\(\dfrac{S_{BDH}}{S_{BCK}}=\left(\dfrac{BH}{BK}\right)^2=\dfrac{BH}{BK}\cdot\dfrac{BD}{BC}=\dfrac{BH}{BC}\cdot\dfrac{BD}{BK}\)

\(=\dfrac{1}{4}\cdot\dfrac{BD\cdot BK}{BK^2}=\dfrac{1}{4}\cdot\dfrac{BA^2}{BK^2}=\dfrac{1}{4}\cdot cos^2ABD\)

=>\(S_{BDH}=\dfrac{1}{4}\cdot S_{BCK}\cdot cos^2ABD\)