Question

Cho tam giác nhọn ABC , góc A=30° . 2 đường cao BH và CK . Chứng minh rằng S AHK = 3S BCHK

Answer 1

TG ABH ∼ TG ACK (g.g) ⇒\(\frac{AH}{AK}=\frac{AB}{AC}\) ⇒\(\frac{AH}{AB}=\frac{AK}{AC}\)

⇒ TG AHK∼ TG ABC (c.g.c )

⇒ \(\frac{S_{AHK}}{S_{ABC}}=\left(\frac{AH}{AB}\right)^2=cos^2A\Rightarrow S_{AHK}=S_{ABC}.cos^2A=S_{ABC}.\left(\frac{\sqrt{3}}{2}\right)^2=\frac{3}{4}S_{ABC}\circledast\)

\(S_{BCHK}=S_{ABC}-S_{AHK}=S_{ABC}-\frac{3}{4}S_{ABC}=\frac{1}{4}S_{ABC}\otimes\)

Từ\(\circledast và\otimes\) ⇒ \(S_{AHK}=3S_{BCHK}\)