Question

Answer 1

a: Xét ΔACH có C'H'//CH

nên \(\frac{AC^{\prime}}{AC}=\frac{AH^{\prime}}{AH}\) (1)

Xét ΔACB có C'B'//CB

nên \(\frac{AC^{\prime}}{AC}=\frac{C^{\prime}B^{\prime}}{CB}\) (2)

Từ (1),(2) suy ra \(\frac{AH^{\prime}}{AH}=\frac{B^{\prime}C^{\prime}}{BC}\)

b: \(AH^{\prime}=\frac{AH}{3}\)

=>\(\frac{AH^{\prime}}{AH}=\frac13\)

=>\(\frac{B^{\prime}C^{\prime}}{BC}=\frac13\)

ΔABC có C'B'//CB

nên ΔAC'B'~ΔACB

=>\(\frac{S_{AC^{\prime}B^{\prime}}}{S_{ACB}}=\left(\frac{C^{\prime}B^{\prime}}{CB}\right)^2=\frac19\)

=>\(S_{AC^{\prime}B^{\prime}}=\frac19\cdot67,5=7,5\left(\operatorname{cm}^2\right)\)