Question

ΔABC, trên AB, AC lấy M, N sao cho \(\dfrac{AM}{MB}\text{=}\dfrac{2}{5}\) ; \(\dfrac{BN}{NC}\text{=}\dfrac{1}{3}\) ; I là giao điểm AN, CM. Tính tỉ số \(\dfrac{AI}{AN};\dfrac{CI}{IM}\)

Answer 1

Vì \(\frac{MA}{MB}=\frac25\)

nên \(S_{CMA}=\frac25\times S_{CMB};S_{IMA}=\frac25\times S_{IMB}\)

=>\(S_{CMA}-S_{IMA}=\frac25\times\left(S_{CMB}-S_{IMB}\right)\)

=>\(S_{CIA}=\frac25\times S_{CIB}\)

Vì \(\frac{BN}{NC}=\frac13\)

nên \(S_{ANB}=\frac13\times S_{ANC};S_{INB}=\frac13\times S_{INC}\)

=>\(S_{ANB}-S_{INB}=\frac13\times\left(S_{ANC}-S_{INC}\right)\)

=>\(S_{AIB}=\frac13\times S_{AIC}\)

=>\(S_{AIB}=\frac13\times\frac25\times S_{CIB}=\frac{2}{15}\times S_{CIB}\)

Vì \(\frac{BN}{NC}=\frac13\)

nên NC=3BN

Ta có: NC+BN=BC

=>BC=3BN+BN=4BN

=>\(BN=\frac14\cdot BC\)

=>\(S_{INB}=\frac14\cdot S_{IBC}\)

=>\(\frac{S_{AIB}}{S_{IBN}}=\frac{2}{15}:\frac14=\frac{8}{15}\)

=>\(\frac{AI}{IN}=\frac{8}{15}\)

=>\(\frac{AI}{AN}=\frac{8}{23}\)

Vì \(\frac{AM}{MB}=\frac25\)

nên \(AM=\frac25MB\)

Ta có: AM+MB=AB

=>\(AB=\frac25MB+MB=\frac75MB\)

=>\(MB=\frac57BA\)

=>\(S_{BMI}=\frac57\cdot S_{BIA}=\frac57\cdot\frac{2}{15}\cdot S_{BIC}=\frac{10}{105}\cdot S_{BIC}=\frac{2}{21}\cdot S_{BIC}\)

=>\(\frac{MI}{IC}=\frac{2}{21}\)

=>\(\frac{CI}{IM}=\frac{21}{2}\)