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八年级数学解答题一般
题目
已知等腰ABC\triangle ABC中,BDACBD\bot AC,且BD=12ACBD=\frac{1}{2}AC,则等腰ABC\triangle ABC的顶角度数为____.
知识点:三角形章节:第13章 三角形 / 13.1 三角形的概念

答案与解析

答案

如图11中,当AB=ACAB=AC时,

BDAC\because BD\bot ACBD=12ACBD=\frac{1}{2}AC
AB=2BD\therefore AB=2BD
A=30\therefore \angle A=30^{\circ}
如图22中,当AB=ACAB=AC

BDAC\because BD\bot ACBD=12ACBD=\frac{1}{2}AC
AB=2BD\therefore AB=2BD
DAB=30\therefore \angle DAB=30^{\circ}
BAC=150\therefore \angle BAC=150^{\circ}

如图33中,当BA=BCBA=BC

BDAC\because BD\bot ACBA=BCBA=BC
BD=AD=DC\therefore BD=AD=DC
A=ABD=CBD=C=45\therefore \angle A=\angle ABD=\angle CBD=\angle C=45^{\circ}
ABC=90\therefore \angle ABC=90^{\circ}
综上所述,满足条件的等腰三角形的顶角的度数为3030^{\circ}150150^{\circ}9090^{\circ}.
故答案为:3030^{\circ}150150^{\circ}9090^{\circ}.

解析

如图11中,当AB=ACAB=AC时,

BDAC\because BD\bot ACBD=12ACBD=\frac{1}{2}AC
AB=2BD\therefore AB=2BD
A=30\therefore \angle A=30^{\circ}
如图22中,当AB=ACAB=AC

BDAC\because BD\bot ACBD=12ACBD=\frac{1}{2}AC
AB=2BD\therefore AB=2BD
DAB=30\therefore \angle DAB=30^{\circ}
BAC=150\therefore \angle BAC=150^{\circ}

如图33中,当BA=BCBA=BC

BDAC\because BD\bot ACBA=BCBA=BC
BD=AD=DC\therefore BD=AD=DC
A=ABD=CBD=C=45\therefore \angle A=\angle ABD=\angle CBD=\angle C=45^{\circ}
ABC=90\therefore \angle ABC=90^{\circ}
综上所述,满足条件的等腰三角形的顶角的度数为3030^{\circ}150150^{\circ}9090^{\circ}.
故答案为:3030^{\circ}150150^{\circ}9090^{\circ}.

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