《2025年广东名师讲练通九年级数学全一册北师大版深圳专版》

3. 如图，$CD = 2BC$，$ED = 2AC$，$BC // DE$，点$A$，$C$，$D$在同一条直线上. 求证：$\triangle ABC \backsim \triangle ECD$.
证明: $\because BC// DE$,
$\therefore ∠ACB = ∠CDE$.
$\because CD = 2BC$, $ED = 2AC$,
$\therefore BC:CD = 1:2$, $AC:ED = 1:2$,
$\therefore \frac{BC}{CD}=\frac{AC}{ED}$,
$\therefore △ABC\backsim △ECD$.

答案： 证明: $\because BC// DE$,
$\therefore ∠ACB = ∠CDE$.
$\because CD = 2BC$, $ED = 2AC$,
$\therefore BC:CD = 1:2$, $AC:ED = 1:2$,
$\therefore △ABC\backsim △ECD$.

4. 如图，已知$∠DAB = ∠ECB$，$∠ABD = ∠CBE$. 求证：$\triangle ABC \backsim \triangle DBE$.
证明: $\because ∠DAB = ∠ECB$, $∠ABD = ∠CBE$,
$\therefore$$\triangle ABD\backsim \triangle CBE$，$\therefore$$\frac{AB}{CB} = \frac{BD}{BE}$，即$\frac{AB}{BD} = \frac{BC}{BE}$.
$\because ∠ABC = ∠ABD + ∠DBC$, $∠DBE = ∠DBC + ∠CBE$,
$\therefore$$∠ABC = ∠DBE$，$\therefore \triangle ABC\backsim \triangle DBE$.

答案： 证明: $\because ∠DAB = ∠ECB$, $∠ABD = ∠CBE$,
$\therefore △ABD\backsim △CBE$, $\therefore \frac{AB}{CB} = \frac{BD}{BE}$, 即 $\frac{AB}{BD} = \frac{BC}{BE}$.
$\because ∠ABC = ∠ABD + ∠DBC$, $∠DBE = ∠DBC + ∠CBE$,
$\therefore ∠ABC = ∠DBE$, $\therefore △ABC\backsim △DBE$.

5. 如图，$D$，$E$分别为$AB$，$AC$边上两点，且$AD = 5$，$BD = 3$，$AE = 4$，$CE = 6$. 求证：$\triangle ADE \backsim \triangle ACB$.
证明: $\because AD = 5$, $BD = 3$, $AE = 4$, $CE = 6$,
$\therefore AB = AD + BD =$8, $AC = AE + CE =$10,
$\therefore \frac{AE}{AB} = \frac{4}{8} =$$\frac{1}{2}$, $\frac{AD}{AC} = \frac{5}{10} =$$\frac{1}{2}$, $\therefore \frac{AE}{AB} = \frac{AD}{AC}$.
又 $∠A = ∠A$, $\therefore △ADE\backsim △ACB$.

答案： 证明: $\because AD = 5$, $BD = 3$, $AE = 4$, $CE = 6$,
$\therefore AB = AD + BD = 8$, $AC = AE + CE = 10$,
$\therefore \frac{AE}{AB} = \frac{4}{8} = \frac{1}{2}$, $\frac{AD}{AC} = \frac{5}{10} = \frac{1}{2}$, $\therefore \frac{AE}{AB} = \frac{AD}{AC}$.
又 $∠A = ∠A$, $\therefore △ADE\backsim △ACB$.

6. 如图，点$E$，$F$分别在正方形$ABCD$的边$BC$，$CD$上，$BE = 3$，$EC = 6$，$CF = 2$. 求证：$\triangle ABE \backsim \triangle ECF$.

答案： 证明: $\because BE = 3$, $EC = 6$, $CF = 2$, $\therefore BC = 3 + 6 = 9$.
$\because$ 四边形 $ABCD$ 是正方形, $\therefore AB = BC = 9$, $∠B = ∠C = 90°$.
$\because \frac{AB}{CE} = \frac{9}{6} = \frac{3}{2}$, $\frac{BE}{CF} = \frac{3}{2}$, $\therefore \frac{AB}{CE} = \frac{BE}{CF}$, $\therefore △ABE\backsim △ECF$.

7. 如图，在$\triangle ABC$中，$∠C = 90^{\circ}$，$BC = 8cm$，$AC = 6cm$，点$Q$从点$B$出发，沿$BC$方向以$2cm/s$的速度移动，点$P$从点$C$出发，沿$CA$方向以$1cm/s$的速度移动，若点$Q$，$P$分别同时从点$B$，$C$出发，试探究经过多少秒后，以点$C$，$P$，$Q$为顶点的三角形与$\triangle ABC$相似？

经过$\frac{12}{5}$秒或$\frac{32}{11}$秒后，以点$C$，$P$，$Q$为顶点的三角形与$\triangle ABC$相似.

答案： 解: 设经过 $x$ 秒后, 两三角形相似. 由题意, 得 $BQ = 2x$, $CP = x$, $\therefore CQ = 8 - 2x$.
①当 $\frac{CQ}{CB} = \frac{CP}{CA}$ 时, $△PQC\backsim △ABC$,
$\therefore \frac{8 - 2x}{8} = \frac{x}{6}$, 解得 $x = \frac{12}{5}$.
②当 $\frac{CQ}{CA} = \frac{CP}{CB}$ 时, $△PQC\backsim △BAC$,
$\therefore \frac{8 - 2x}{6} = \frac{x}{8}$, 解得 $x = \frac{32}{11}$.
$\therefore$ 经过 $\frac{12}{5}$ 秒或 $\frac{32}{11}$ 秒时, 以点 $C$, $P$, $Q$ 为顶点的三角形与 $△ABC$ 相似.