1. (2024·兰州)如图,在$\triangle ABC$中,$AB=AC,∠BAC=130^{\circ },DA⊥AC$,则$∠ADB=$ (
A.$100^{\circ }$
B.$115^{\circ }$
C.$130^{\circ }$
D.$145^{\circ }$
B
)A.$100^{\circ }$
B.$115^{\circ }$
C.$130^{\circ }$
D.$145^{\circ }$
答案：B
2. 某城市几条道路的位置关系如图所示,道路$AB// CD$,道路$AB$与$AE$的夹角$∠BAE=54^{\circ }$,城市规划部门想新修一条道路$CE$,要求使$CF=EF$,则$∠E$的度数为 (
A.$23^{\circ }$
B.$25^{\circ }$
C.$27^{\circ }$
D.$30^{\circ }$
C
)A.$23^{\circ }$
B.$25^{\circ }$
C.$27^{\circ }$
D.$30^{\circ }$
答案：C
3. 如图,在$\triangle ABC$中,$BC=5cm$,$BP$,$CP$分别是$∠ABC$和$∠ACB$的平分线,且$PD// AB$,$PE// AC$,则$\triangle PDE$的周长是
5
cm.答案：5
4. (2024·徐州期末)如图,在$\triangle ABC$中,$AB=AC$,$AD⊥BC$,点$E$在$CA$的延长线上,$EF// AD$.求证:$AE=AF$.
证明：
证明：
$\because AB = AC$，$AD \perp BC$，$\therefore \angle BAD = \angle DAC$。$\because EF // AD$，$\therefore \angle BAD = \angle AFE$，$\angle DAC = \angle E$，$\therefore \angle E = \angle AFE$，$\therefore AE = AF$。
答案：证明：$\because AB = AC$，$AD \perp BC$，
$\therefore \angle BAD = \angle DAC$。$\because EF // AD$，
$\therefore \angle BAD = \angle AFE$，$\angle DAC = \angle E$，
$\therefore \angle E = \angle AFE$，$\therefore AE = AF$。
$\therefore \angle BAD = \angle DAC$。$\because EF // AD$，
$\therefore \angle BAD = \angle AFE$，$\angle DAC = \angle E$，
$\therefore \angle E = \angle AFE$，$\therefore AE = AF$。
5. 如图,在$\triangle ABC$中,$AD⊥BC$,$∠B=62^{\circ }$,$AB+BD=CD$,则$∠BAC$的度数为 (
A.$87^{\circ }$
B.$88^{\circ }$
C.$89^{\circ }$
D.$90^{\circ }$
A
)A.$87^{\circ }$
B.$88^{\circ }$
C.$89^{\circ }$
D.$90^{\circ }$
答案：A
6. (2023·如皋月考)如图,在$\triangle ABC$中,$D$为边$AC$上一点,且$BD$平分$∠ABC$,过点$A$作$AE⊥BD$于点$E$.若$∠ABC=52^{\circ }$,$∠C=32^{\circ }$,$AB=5.2$,$BC=9.8$,则$AE=$
2.3
.答案：2.3
7. 如图,$\triangle ABC$的面积为$10cm^{2}$,$AP$垂直于$∠ABC$的平分线$BP$于点$P$,则$\triangle PBC$的面积是
$5cm^{2}$
.答案：$5cm^{2}$