Question

gấp phần c mn ơi

Answer 1

c: CE*CA=CF*CB

=>\(\dfrac{CE}{CB}=\dfrac{CF}{CA}\)

Xét ΔCEF vuông tại C và ΔCBA vuông tại C có

\(\dfrac{CE}{CB}=\dfrac{CF}{CA}\)

Do đó: ΔCEF đồng dạng với ΔCBA

=>\(\dfrac{S_{CBA}}{S_{CEF}}=\left(\dfrac{AB}{EF}\right)^2=\left(\dfrac{AB}{CH}\right)^2\)

\(\dfrac{1}{sin^2CAB}+\dfrac{1}{cos^2HCB}\)

\(=\dfrac{1}{sin^2CAB}+\dfrac{1}{cos^2CAB}\)

\(=\dfrac{cos^2CAB+sin^2CAB}{\left(sinCAB\cdot cosCAB\right)^2}=\dfrac{1}{\left(\dfrac{CB}{BA}\cdot\dfrac{CA}{BA}\right)^2}\)

\(=\dfrac{1}{\left(\dfrac{CH\cdot AB}{AB^2}\right)^2}=\dfrac{1}{\left(\dfrac{CH}{AB}\right)^2}=\dfrac{AB^2}{CH^2}\)

=>\(\dfrac{S_{CBA}}{S_{CEF}}=\dfrac{1}{sin^2CAB}+\dfrac{1}{cos^2HCB}\)(ĐPCM)