《2025年5年中考3年模拟数学广东专版》

变式1 如图，△ABC中，AB = AC，AD⊥BC于D点，DE⊥AB于点E，BF⊥AC于点F，DE = 3 cm，则BF = ________cm.

答案： 答案 6
解析 $\because AB = AC$,$\therefore \angle C=\angle ABC$,
又$\because AD\perp BC$于$D$点,
$\therefore BD = DC=\frac{1}{2}BC$,
又$DE\perp AB,BF\perp AC$,
$\therefore \angle BED=\angle CFB = 90^{\circ}$,
$\therefore \triangle BED\sim\triangle CFB$,
$\therefore DE:BF = BD:BC = 1:2$,
$\therefore BF = 2DE = 2\times3 = 6$ cm.

例2 如图，点A、D、C、F在一条直线上，且AD = CF，AB = DE，∠BAC = ∠EDF.求证：∠B = ∠E.

解题关键点 由AD = CF，可得AC = DF.

答案： 证明 $\because AD = CF$,
$\therefore AD + CD = CF + CD$,$\therefore AC = DF$,
$\because$ 在$\triangle ABC$和$\triangle DEF$中,
$\begin{cases}AB = DE,\\\angle BAC=\angle EDF,\\AC = DF,\end{cases}$
$\therefore \triangle ABC\cong\triangle DEF(SAS)$,
$\therefore \angle B=\angle E$.

变式2 如图，在△ABC中，D是边BC上的点，DE⊥AC，DF⊥AB，垂足分别为E，F，且DE = DF，CE = BF.求证：∠B = ∠C.

答案： 证明 $\because DE\perp AC,DF\perp AB$,
$\therefore \angle DEC=\angle DFB = 90^{\circ}$.
在$\triangle DEC$和$\triangle DFB$中,
$\begin{cases}DE = DF,\\\angle DEC=\angle DFB,\\CE = BF,\end{cases}$
$\therefore \triangle DEC\cong\triangle DFB(SAS)$,
$\therefore \angle B=\angle C$.

变式3 如图，∠BAC = 90°，AD是∠BAC内部一条射线，若AB = AC，BE⊥AD于点E，CF⊥AD于点F.求证：AF = BE.

答案： 证明 $\because BE\perp AD,CF\perp AD$
$\therefore \angle BEA=\angle AFC = 90^{\circ}$,
$\because \angle BAE+\angle CAF = 90^{\circ},\angle C+\angle CAF = 90^{\circ}$,$\therefore \angle BAE=\angle C$,
在$\triangle ABE$和$\triangle CAF$中,
$\begin{cases}\angle BAE=\angle C,\\\angle BEA=\angle AFC,\\AB = CA,\end{cases}$
$\therefore \triangle ABE\cong\triangle CAF(AAS)$,
$\therefore AF = BE$.