《2025年长江全能学案同步练习册九年级数学全一册人教版》

例1 下列条件,能判定$\triangle ABC \backsim \triangle A'B'C'$的是(C)
A.$\frac{AB}{A'B'} = \frac{AC}{A'C'}$
B.$\frac{AB}{AC} = \frac{A'B'}{A'C'}且\angle A = \angle C'$
C.$\frac{AB}{A'B'} = \frac{BC}{B'C'}且\angle B = \angle B'$
D.$\frac{AB}{A'B'} = \frac{AC}{A'C'}且\angle B = \angle B'$
【思路导析】画图按定理判定.
【请你解答】______.

答案： C

例2 如图,$BD平分\angle ABC$,$AB = 4$,$BC = 9$,当$BD = $6时,$\triangle ABD \backsim \triangle DBC$.

【思路导析】当$\frac{AB}{BD} = \frac{BD}{BC}$时,$\triangle ABD \backsim \triangle DBC$.
【请你解答】6.

答案： 6

例3 如图,已知$P是正方形ABCD的边BC$上的一点,且$BP = 3PC$,$Q是DC$的中点,求证：$AQ \perp PQ$.

答案：
【探究点拨】证明$\angle AQD = \angle QPC$可以达到目的,可从证明$\triangle ADQ \backsim \triangle QCP$入手.

【规范解答】$\because ABCD$是正方形,$\therefore \angle C = \angle D = 90^{\circ}$.
设$PC = a$,则$BP = 3a$,$BC = CD = AD = 4a$.
$\because Q是DC$的中点,$\therefore CQ = DQ = 2a$.
$\therefore \frac{PC}{DQ} = \frac{a}{2a} = \frac{1}{2}$,$\frac{CQ}{AD} = \frac{2a}{4a} = \frac{1}{2}$,
$\therefore \frac{PC}{DQ} = \frac{CQ}{AD}$,而$\angle C = \angle D$.
$\therefore \triangle ADQ \backsim \triangle QCP$,$\therefore \angle AQD = \angle QPC$.
$\because \angle PQC + \angle QPC = 90^{\circ}$,
$\therefore \angle PQC + \angle AQD = 90^{\circ}$,
$\therefore \angle AQP = 180^{\circ} - 90^{\circ} = 90^{\circ}$,
即$AQ \perp PQ$.

1. 如图,$E是四边形ABCD的对角线BD$上的一点,且$\frac{AB}{AE} = \frac{AC}{AD}$,$\angle 1 = \angle 2$,求证：$\angle ABC = \angle AED$.

答案：
∵∠1=∠2,
∴∠1+∠EAC=∠2+∠EAC.即∠BAC=∠EAD.又$\frac{AB}{AE}=\frac{AC}{AD}$,
∴△ABC∽△AED.
∴∠ABC=∠AED.

2. 如图,已知$AD \cdot AC = AB \cdot AE$,求证：$\triangle ADE \backsim \triangle ABC$.

答案：
∵AD·AC=AE·AB,
∴$\frac{AD}{AB}=\frac{AE}{AC}$.在△ADE 和△ABC 中，
∵$\frac{AD}{AB}=\frac{AE}{AC}$，∠A=∠A，
∴△ADE ∽△ABC.