Cho tam giác ABC vuông tại A, đường cao AH. Biết BC = 8cm, BH = 2cm.

Trả lời

a) Áp dụng HTL tam giác:

$\left\{\begin{array}{c}A{B}^2=BH.BC=16\\ A{C}^2=BC.CH=8\left(8-2\right)=48\\ A{H}^2=BH.CH=2\left(8-2\right)=12\end{array}\right.\Rightarrow \left\{\begin{array}{c}AB=4\left(cm\right)\\ AC=4\sqrt{3}\left(cm\right)\\ AH=2\sqrt{3}\left(cm\right)\end{array}\right.$

b) $\widehat{ADB}=\widehat{AHB}\left(=90°\right)\Rightarrow$ADHB nội tiếp

$\Rightarrow \widehat{DHA}=\widehat{DBA}$ (cùng chắn AD) (1)

$\left\{\begin{array}{c}\widehat{CKB}=\widehat{KAB}+\widehat{ABD}=90°+\widehat{ABD}\\ \widehat{DHB}=\widehat{DHA}+\widehat{AHB}=\widehat{DHA}+90°\\ \widehat{ABD}=\widehat{DHA}\left(cmt\right)\end{array}\right.$

$\Rightarrow \widehat{CKB}=\widehat{DHB}$

$\left\{\begin{array}{c}\widehat{CKB}=\widehat{DHB}\\ \widehat{CBK}chung\end{array}\right.\Rightarrow \Delta DHB~\Delta CKB\left(g.g\right)$

$\Rightarrow \frac{BD}{BC}=\frac{BH}{BK}\Rightarrow BD.BK=BH.BC$

c) Áp dụng công thức tính diện tích hình tam giác bằng nửa tích hai cạnh nhân sin góc xen giữa

$S_{BHD}=\frac{1}{2}BH.BD.\sin \widehat{DBH}$

$S_{BKC}=\frac{1}{2}BK.BC.\sin \widehat{KBC}$

Mà $\widehat{DBH}=\widehat{KBC}$

$\Rightarrow \frac{S_{BHD}}{S_{BKC}}=\frac{BH.BD}{BK.BC}=\frac{2BD}{8BK}=\frac{BD}{4BK}=\frac{B{D}^2}{4BK.BD}$

$=\frac{1}{4}\frac{B{D}^2}{A{B}^2}$ (hệ thức lượng) $=\frac{1}{4}.{\cos }^2\widehat{ABD}$

$\Rightarrow S_{BHD}=\frac{1}{4}{S}_{BKC}.{\cos }^2\widehat{ABD}$.