toán 8 khó đây ^^

N

nhuquynhdat

Bài 1

a) xét $\Delta ABD$ có $NO//AB \to \dfrac{NO}{AB}= \dfrac{OD}{BD}$

Tương tự ta có:

$\dfrac{MO}{AB}= \dfrac{OC}{AC}$

$\dfrac{NO}{CD}= \dfrac{AO}{AC}$

$\dfrac{MO}{CD}= \dfrac{OB}{BD}$

$\to \dfrac{NO}{AB}+ \dfrac{MO}{AB}+ \dfrac{NO}{CD}+\dfrac{MO}{CD}= \dfrac{OD}{BD}+ \dfrac{OC}{AC}+ \dfrac{AO}{AC}+ \dfrac{OB}{BD}$

$\to \dfrac{MN}{AB}+ \dfrac{MN}{CD}= 2$

$\to \dfrac{1}{AB}+ \dfrac{1}{CD}= \dfrac{2}{MN}$

b) Kẻ $AH \perp DC ; BK \perp AC$

$S_{ADC}= S_{BDC}$ ( Tự CM)

$\to S_{AOD}=S_{BOC}$

Ta có:

$S_{AOB}= \dfrac{1}{2}BK.OA$

$S_{COB}= \dfrac{1}{2}BK.OC$

$\to \dfrac{S_{AOB}}{S_{BOC}}= \dfrac{OA}{OC}$

Hoàn toàn tương tự có:

$\dfrac{S_{AOD}}{S_{DOC}}= \dfrac{OA}{OC}$

$\to \dfrac{S_{AOB}}{S_{BOC}}= \dfrac{S_{AOD}}{S_{DOC}}$

$\to S_{AOB}.S_{DOC}= S_{BOC}.S_{AOD}$

$\to a^2.b^2 = S^2_{BOC}$

$\to S_{BOC}=S_{AOD}= |a|.|b|=ab$

$S_{ABCD}=S_{ABO}+ S_{AOD}+ S_{DOC}+ S_{BOC}$

$=S_{ABO}+ 2S_{AOD}+ S_{DOC}= a^2+ 2ab+ b^2=(a+b)^2$
 
Last edited by a moderator:
You must log in or register to reply here.
Top Bottom