1.(教材变式)在$\triangle ABC$中,$∠B= ∠A-10^{\circ },∠C= ∠B+20^{\circ }$,求$∠B$的度数.

解:设$\angle B=x$.
$\because \angle B=\angle A-10^{\circ}$,$\angle C=\angle B+20^{\circ}$,
$\therefore \angle A=\angle B+10^{\circ}=x+10^{\circ}$,
$\angle C=x+20^{\circ}$.
$\because \angle A+\angle B+\angle C=180^{\circ}$,
$\therefore x+10^{\circ}+x+x+20^{\circ}=180^{\circ}$,
解得$x=50^{\circ}$,
$\therefore \angle B=50^{\circ}$.

2.(教材变式)如图,在$\triangle ABC$中,$∠ACB= ∠ABC= \frac {1}{4}∠BAC$,CD 是 AB 边上的高.求$∠DCA$的度数.

解:设$\angle ABC=\angle ACB=x$,
则$\angle BAC=4x$,
$\therefore x+x+4x=180^{\circ}$,
$\therefore x=30^{\circ}$,
$\therefore \angle DCB=90^{\circ}-\angle B=60^{\circ}$,
$\therefore \angle DCA=\angle DCB-\angle ACB$
$=60^{\circ}-30^{\circ}$
$=30^{\circ}$.

3.如图,AD 平分$∠BAC,∠EAD= ∠EDA,∠B= 50^{\circ },∠CAD:∠E= 1:3$.求$∠E$的度数.

解:设$\angle CAD=x$,则$\angle E=3x$.
$\because AD$平分$\angle BAC$,
$\therefore \angle BAD=\angle CAD=x$,
$\therefore \angle EAD=\angle ADE$
$=\angle B+\angle BAD$
$=50^{\circ}+x$.
$\because$在$\triangle ADE$中,
$\angle E+\angle ADE+\angle DAE=180^{\circ}$,
$\therefore 3x+2(50^{\circ}+x)=180^{\circ}$,
解得$x=16^{\circ}$,
$\therefore \angle E=3x=48^{\circ}$.

4.如图,在$\triangle ABC$中,D 为 BC 上一点,$BE⊥AC$于点 E,交 AD 于点 F,$∠ABD= ∠ADB,∠DAC= ∠C= 2∠BAD$.求$∠BFD$的度数.

解:设$\angle BAD=x$,
则$\angle DAC=\angle C=2\angle BAD=2x$,
$\angle ABD=\angle ADB=4x$.
$\because \angle ABD+\angle ADB+\angle BAD=180^{\circ}$,
$\therefore 4x+4x+x=180^{\circ}$,解得$x=20^{\circ}$,
$\therefore \angle DAC=2x=40^{\circ}$.
$\because BE\perp AC$,
$\therefore \angle AEF=90^{\circ}$,
$\therefore \angle AFE=90^{\circ}-\angle DAC$
$=90^{\circ}-40$
$=50^{\circ}$,
$\therefore \angle BFD=\angle AFE=50^{\circ}$.