[Hình 9] Chứng minh

E

eye_smile

Ta có:

$S_{BKHC}=S_{ABC}-S_{AKH}=\dfrac{1}{2}.sinA.AB.AC-\dfrac{1}{2}.sinA.AK.AH=\dfrac{1}{2}.sinA(AB.AC-AK.AH)$

Ta cần cm: $AB.AC-AK.AH=CK.BH$

$\Delta AKC$ đ.dạng với $\Delta AHB$

\Rightarrow $\dfrac{AK}{AH}=\dfrac{AC}{AB}=\dfrac{CK}{BH}$

\Rightarrow $AB.AC-AK.AH=AC.\dfrac{AC.AH}{AK}-AK.AH=\dfrac{AH}{AK}(AC^2-AK^2)=\dfrac{AH}{AK}.CK^2=CK.\dfrac{AH}{AK}.\dfrac{AK.BH}{AH}=CK.BH$

\Rightarrow đpcm.
 
You must log in or register to reply here.
Top Bottom