第103页 - 通城学典课时作业本九年级数学人教版南通专版

$\frac{1}{9}$

(0,2)

$\frac{4\sqrt{2}}{5}$

$\frac{12\sqrt{5}}{11} $

$解:(2)S_{△ABC}=\frac{1}{2}×1×2=1$

$(3)∵△ABC与△AB'C'的相似比为\frac{1}{3}$

$∴\frac{S_{△ABC}}{S_{△AB'C'}}=\frac{1}{9}$

$∴S_{△AB'C'}=9\ $

$证明:(1)∵AB=AC,∴∠B=∠C$

$∵∠BDE=180°-∠B-∠DEB,∠CEF=180°-∠DEF-∠DEB$

$又∵∠DEF=∠B,∴∠BDE=∠CEF,∴△BDE∽△CEF,∴\frac{BE}{CF}=\frac{DE}{EF}$

$(2)由(1)，得\frac{BE}{CF}=\frac{DE}{EF}$

$∵ E是BC的中点，∴ BE=CE,∴ \frac{CE}{CF}=\frac{DE}{EF},即\frac{DE}{EC}=\frac{EF}{CF}$

$∵ AB=AC，∴ ∠C=∠B=∠DEF,∴△DEF∽△ECF$

$∴∠DFE=∠EFC,∴FE平分∠DFC$