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S(KCD) = CD x BC X 1/2 = 1/2 S(ABCD)
-S(ABNM) = S(CDMN) = 1/2 s(ABCD) ( Vì AM = NC, DM = BN, AB = CD)
=> S(ABNM) = S(KCD)
=> S(CDEF) = S(AKEM) + S(BKFN) ( cùng chung S(KEF)
- Mà S(ABNM) = S(CDMN) => S(KEF) = S(DME) + S(CNF) ( cùng bớt S(CDEF) = S(AKEM) + S(BKFN))
-S(KCD) = CD x BC X 1/2 = 1/2 S(ABCD) -S(ABNM) = S(CDMN) = 1/2 s(ABCD) ( Vì AM = NC, DM = BN, AB = CD) => S(ABNM) = S(KCD) => S(CDEF) = S(AKEM) + S(BKFN) ( cùng chung S(KEF) - Mà S(ABNM) = S(CDMN) => S(KEF) = S(DME) + S(CNF) ( cùng bớt S(CDEF) = S(AKEM) + S(BKFN))
Mình giải theo cách lớp 5.
a) Có: \(AN+NC=AC\) mà \(AN=\dfrac{1}{2}NC\)
\(\Rightarrow\dfrac{1}{2}NC+NC=AC\Rightarrow\dfrac{3}{2}NC=AC\Rightarrow NC=\dfrac{2}{3}AC\)
\(2AN=\dfrac{2}{3}AC\Rightarrow AN=\dfrac{2}{3}.\dfrac{1}{2}AC=\dfrac{1}{3}AC\)
\(\dfrac{S_{ABN}}{S_{ABC}}=\dfrac{AN}{AC}=\dfrac{1}{3}\Rightarrow S_{ABN}=\dfrac{1}{3}S_{ABC}\left(1\right)\)
\(\dfrac{S_{ACM}}{S_{ABC}}=\dfrac{AM}{AB}=\dfrac{1}{3}\Rightarrow S_{ACM}=\dfrac{1}{3}S_{ABC}\left(2\right)\)
Từ (1) và (2) suy ra:
\(S_{ABN}=S_{ACM}\)
\(\Rightarrow S_{ABN}-S_{AMON}=S_{ACM}-S_{AMON}\)
\(\Rightarrow S_{MOB}=S_{NOC}\).
b) \(\dfrac{S_{AMC}}{S_{AMN}}=\dfrac{AC}{AN}=3\Rightarrow S_{AMC}=3S_{AMN}=3.4,5=13,5\left(cm^2\right)\)
\(\dfrac{S_{ABC}}{S_{AMN}}=\dfrac{AB}{AM}=3\Rightarrow S_{ABC}=3S_{AMN}=3.13,5=40,5\left(cm^2\right)\)
\(\dfrac{S_{NCB}}{S_{ABC}}=\dfrac{NC}{AC}=\dfrac{2}{3}\Rightarrow S_{NCB}=\dfrac{2}{3}S_{ABC}=\dfrac{2}{3}.40,5=27\left(cm^2\right)\)